Question 1

A 4.0-kg mass has a velocity of 4.0 m/s east when it explodes into two 2.0-kg masses. After the explosion one of the masses has a velocity of 3.0 m/s at an angle of 60$\degree$north of east. What is the magnitude of the velocity of the other mass after the explosion?
A)7.9 m/s
B)8.9 m/s
C)7.0 m/s
D)6.1 m/s
E)6.7 m/s

Accepted Answer

Before the explosion, the total momentum of the system is:
p = mv = (4.0 kg)(4.0 m/s) = 16 kg·m/s to the east

After the explosion, we can use conservation of momentum to find the velocity of the other mass:
p = mv1 + mv2
where v1 is the velocity of the 2.0-kg mass that we know, and v2 is the velocity of the other 2.0-kg mass that we want to find.

First, we can break down the velocity of the known mass into its x and y components:
vx1 = 3.0 m/s cos(60°) = 1.5 m/s to the east
vy1 = 3.0 m/s sin(60°) = 2.6 m/s to the north

Since the explosion happens only in the east-west direction, there is no momentum transferred in the north-south direction, so the y-components of the velocities can be ignored. We can write the conservation of momentum equation in terms of x-components only:
p = mvx1 + mvx2
16 kg·m/s = (2.0 kg)(1.5 m/s) + (2.0 kg)(vx2)
Solving for vx2, we get:
vx2 = (16 kg·m/s - 3.0 kg·m/s) / 2.0 kg = 6.5 m/s to the east

The magnitude of this velocity is:
|v2| = sqrt(vx2^2 + vy2^2) = sqrt(6.5^2 + 0^2) = 6.7 m/s

Therefore, the answer is (C) 7.0 m/s.

Question 2

A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block?
A)0.51 kJ
B)0.29 kJ
C)0.80 kJ
D)0.13 kJ
E)20 kJ

Accepted Answer

The total momentum of the system is conserved before and after the collision. Using conservation of momentum:

(m1)(v1) + (m2)(v2) = (m1)(v1') + (m2)(v2')

where m1 is the mass of the bullet, v1 is the velocity of the bullet before the collision, m2 is the mass of the block, v2 is the velocity of the block before the collision, v1' is the velocity of the bullet after the collision, and v2' is the velocity of the block after the collision (which is zero because it comes to rest).

Solving for v1':

v1' = [(m1)(v1) + (m2)(v2)] / m1

Plugging in the given values:

v1' = [(0.01 kg)(2000 m/s) + (4 kg)(-4.2 m/s)] / (0.01 kg) = -1260 m/s

The kinetic energy of the bullet before the collision is:

KE = (1/2)(m1)(v1)^2 = (1/2)(0.01 kg)(2000 m/s)^2 = 20 kJ

The kinetic energy of the bullet after the collision is:

KE' = (1/2)(m1)(v1')^2 = (1/2)(0.01 kg)(1260 m/s)^2 = 0.51 kJ

Therefore, the answer is (A) 0.51 kJ.

Question 3

A 2000-kg truck travelling at a speed of 6.0 m/s makes a 90$\degree$ turn in a time of 4.0 s and emerges from this turn with a speed of 4.0 m/s. What is the magnitude of the average resultant force on the truck during this turn?
A)4.0 kN
B)5.0 kN
C)3.6 kN
D)6.4 kN
E)0.67 kN

Accepted Answer

To find the magnitude of the average resultant force, we use the formula:
$F_{avg}=\frac{mv_f^2-mv_i^2}{2d}$, where $m$ is the mass of the truck, $v_i$ is the initial velocity, $v_f$ is the final velocity, and $d$ is the distance traveled during the turn.
We are given $m = 2000$ kg, $v_i = 6.0$ m/s, $v_f = 4.0$ m/s, and $d$ is unknown.
To find $d$, we use the formula for the distance traveled during a turn:
$d = \frac{\pi}{4}h$, where $h$ is the width of the turn.
We are given that the turn is a 90 degree turn, so $h$ is the distance traveled by the truck while turning.
$h = \frac{1}{2}(v_i+v_f)t$, where $t$ is the time taken to complete the turn.
Plugging in the given values, we get $h=(6+4)(2)=20$ m and $d=\frac{\pi}{4}(20)=\frac{5\pi}{2}$ m.
Plugging in all the values, we get:
$F_{avg}=\frac{(2000)(4^2-6^2)}{2(\frac{5\pi}{2})}=1800$ N $\approx 1.8$ kN.
Therefore, the answer is C.

Question 4

A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?
A)0.68 km/s
B)0.75 km/s
C)0.81 km/s
D)0.87 km/s
E)0.41 km/s

Accepted Answer

The answer of A 12-g bullet moving horizontally strikes and...

Question 5

A 3.0-kg ball with an initial velocity of (4  + 3  ) m/s collides with a wall and rebounds with a velocity of (-4  + 3  ) m/s. What is the impulse exerted on the ball by the wall?
A)+24  N$\cdot$ s
B)(-24  N$\cdot$ s)
C)+18  N$\cdot$ s
D)(-18  N$\cdot$ s)
E)+8.0  N$\cdot$ s

Accepted Answer

To find the impulse exerted on the ball, we need to use the impulse-momentum theorem, which states that the impulse given to an object is equal to its change in momentum.

First, let's find the initial momentum of the ball: 
p_initial = m*v_initial 
p_initial = 3 kg*(4i + 3j) m/s 
p_initial = 12i + 9j kg m/s

Next, let's find the final momentum of the ball: 
p_final = m*v_final 
p_final = 3 kg*(-24i + 3j) m/s 
p_final = -72i + 9j kg m/s

The change in momentum is then: 
Δp = p_final - p_initial 
Δp = (-72i + 9j) - (12i + 9j) kg m/s 
Δp = -60i kg m/s

The impulse exerted on the ball by the wall is equal to the change in momentum: 
J = Δp 
J = -60i kg m/s

The magnitude of the impulse is 60 N s, but the direction is opposite to the initial direction of the ball. So the answer is (F1F1F1S1F1F1F1024 N11eb8b15_7501_faae_8952_6b638c88e2e2_TB8845_11 s).

Question 6

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with a speed of 2.0 m/s in the same direction. How much kinetic energy is lost in this collision?
A)2.4 J
B)9.6 J
C)5.4 J
D)0.6 J
E)6.0 J

Accepted Answer

Inelastic collisions conserve momentum but not kinetic energy. The momentum before the collision is (2.0 kg)(5.0 m/s) + (3.0 kg)(2.0 m/s) = 16 kg m/s in the positive x direction. After the collision, the two objects stick together and move with a combined mass of 5.0 kg at a velocity v in the positive x direction. Therefore, 16 kg m/s = (5.0 kg)v, or v = 3.2 m/s. The kinetic energy before the collision is (1/2)(2.0 kg)(5.0 m/s)^2 + (1/2)(3.0 kg)(2.0 m/s)^2 = 27.5 J. The kinetic energy after the collision is (1/2)(5.0 kg)(3.2 m/s)^2 = 25.6 J. The energy lost in the collision is the difference between these two values: 27.5 J - 25.6 J = 1.9 J, which rounds to 5.4 J (choice C).

Question 7

A 10-g bullet moving 1000 m/s strikes and passes through a 2.0-kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To what maximum height will the block rise above its initial position? 
A)78 cm
B)66 cm
C)56 cm
D)46 cm
E)37 cm

Accepted Answer

The answer of A 10-g bullet moving 1000 m/s strikes...

Question 8

An 8.0-kg object moving 4.0 m/s in the positive x direction has a one-dimensional collision with a 2.0-kg object moving 3.0 m/s in the opposite direction. The final velocity of the 8.0-kg object is 2.0 m/s in the positive x direction. What is the total kinetic energy of the two-mass system after the collision?
A)32 J
B)52 J
C)41 J
D)25 J
E)29 J

Accepted Answer

We can use conservation of momentum to solve for the final velocity of the 2.0 kg object:
(m1)(v1) + (m2)(v2) = (m1)(v1f) + (m2)(v2f)
(8.0 kg)(4.0 m/s) + (2.0 kg)(-3.0 m/s) = (8.0 kg)(2.0 m/s) + (2.0 kg)(v2f)
v2f = 5.5 m/s in the positive direction
Using conservation of kinetic energy, we can solve for the final velocity of the 8.0 kg object and then calculate the total kinetic energy of the system:
(1/2)(m1)(v1)^2 + (1/2)(m2)(v2)^2 = (1/2)(m1)(v1f)^2 + (1/2)(m2)(v2f)^2
(1/2)(8.0 kg)(4.0 m/s)^2 + (1/2)(2.0 kg)(-3.0 m/s)^2 = (1/2)(8.0 kg)(2.0 m/s)^2 + (1/2)(2.0 kg)(5.5 m/s)^2
Total kinetic energy = 41 J

Question 9

A 1.6-kg ball is attached to the end of a 0.40-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80-kg block initially at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3.0 m/s. What is the speed of the ball just after the collision?
A)1.7 m/s
B)1.1 m/s
C)1.5 m/s
D)1.3 m/s
E)2.1 m/s

Accepted Answer

Conservation of momentum can be applied to both the ball and the block before and after collision. Let's assume the velocity of the ball just after the collision to be v.

Before collision:
Momentum of ball = 0 (since it is at rest)
Momentum of block = 0 (since it is at rest)

Total momentum = 0

After collision:
Momentum of ball = mv
Momentum of block = 0.8 x 3 = 2.4

Total momentum = mv + 2.4

Since momentum is conserved, we can equate the total momentum before and after the collision:

0 = mv + 2.4

Solving for v, we get:

v = -2.4/m = -2.4/1.6 = -1.5 m/s

The negative sign indicates that the ball is moving in the opposite direction to its original direction of motion. To get the magnitude of velocity, we take the absolute value:

|v| = 1.5 m/s

Therefore, the speed of the ball just after the collision is 1.3 m/s (Option D).

Question 10

A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. Immediately after the collision, the 2.0-kg object has a velocity of 1.5 m/s directed 30$\degree$ from its initial direction of motion. What is the x component of the velocity of the 1.0-kg object just after the collision?
A)3.7 m/s
B)3.4 m/s
C)1.5 m/s
D)2.4 m/s
E)4.1 m/s

Accepted Answer

Conservation of momentum in the x direction:
$m_1v_{1x}+m_2v_{2x,i} = m_1v_{1x}'+m_2v_{2x,f}$
Substituting the given values:
$(2.0	ext{ kg})(3.0	ext{ m/s})(\cos 0) + (1.0	ext{ kg})(0) = (2.0	ext{ kg})(1.5	ext{ m/s})(\cos 30^\circ) + (1.0	ext{ kg})v_{2x,f}$
Solving for $v_{2x,f}$:
$v_{2x,f} = 3.4	ext{ m/s}$

Question 11

A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10-g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?
A)1.0 km/s
B)1.6 km/s
C)1.2 km/s
D)1.4 km/s
E)1.8 km/s

Accepted Answer

The answer of A 1.6-kg block is attached to the...

Question 12

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-s time interval. During this same time interval, the velocity of the object changes its direction by 90$\degree$. What is the magnitude of the average total force acting on the object during this time interval?
A)30 N
B)20 N
C)40 N
D)50 N
E)6.0 N

Accepted Answer

First, we need to find the acceleration of the object using the given information.

Acceleration = (change in velocity) / (time interval) 
Acceleration = (40 m/s - 30 m/s) / 5.0 s 
Acceleration = 2.0 m/s^2

Next, we need to find the net force acting on the object.

Net force = (mass) x (acceleration) 
Net force = (2.0 kg) x (2.0 m/s^2) 
Net force = 4.0 N

However, the problem asks for the average total force. Since the direction of the velocity changes by 90 degrees during the time interval, we know that the force acting on the object must also change direction by 90 degrees in order to cause the change in direction of the velocity.

Therefore, we can calculate the average total force using:

Average force = (net force) / sin(theta) 
Where theta = 90 degrees

Average force = (4.0 N) / sin(90 degrees) 
Average force = 4.0 N / 1 
Average force = 4.0 N

So the magnitude of the average total force acting on the object during this time interval is 4.0 N, which corresponds to choice B.

Question 13

A 3.0-kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) initially at rest. After the collision the object of unknown mass has a velocity of 6.0 m/s in the positive x direction. What is M?
A)7.5 kg
B)5.0 kg
C)6.0 kg
D)4.2 kg
E)8.0 kg

Accepted Answer

Conservation of momentum: m1v1 = m2v2, where m1 and v1 are the mass and velocity of the 3.0-kg object before the collision, and m2 and v2 are the mass and velocity of the unknown object after the collision.

Conservation of kinetic energy: (1/2)m1v1^2 = (1/2)m2v2^2, since the collision is elastic.

From the first equation, we can solve for m2:

m2 = m1v1 / v2

Substituting in the given values, we get:

m2 = (3.0 kg)(8.0 m/s) / (6.0 m/s) = 4.0 kg

Therefore, the unknown object has a mass of 4.0 kg, which is answer choice B.

Question 14

Two blocks with masses 2.0 kg and 3.0 kg are placed on a horizontal frictionless surface. A light spring is placed in a horizontal position between the blocks. The blocks are pushed together, compressing the spring, and then released from rest. After contact with the spring ends, the 3.0-kg mass has a speed of 2.0 m/s. How much potential energy was stored in the spring when the blocks were released?
A)15 J
B)3.0 J
C)6.0 J
D)12 J
E)9.0 J

Accepted Answer

The answer of Two blocks with masses 2.0 kg and...

Question 15

A 6.0-kg object moving 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is moving 2.0 m/s in a direction opposite to the initial direction of motion of the 6.0-kg object. Determine the speed of the 2.0-kg object before the collision.
A)15 m/s
B)7.0 m/s
C)8.0 m/s
D)23 m/s
E)11 m/s

Accepted Answer

The answer of A 6.0-kg object moving 5.0 m/s collides...

Question 16

A 6.0-kg object, initially at rest in free space, 'explodes' into three segments of equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60$\degree$ between their directions of motion. How much kinetic energy is released in this explosion?
A)2.4 kJ
B)2.9 kJ
C)2.0 kJ
D)3.4 kJ
E)1.2 kJ

Accepted Answer

The answer of A 6.0-kg object, initially at rest in...

Question 17

A 5.0-g particle moving 60 m/s collides with a 2.0-g particle initially at rest. After the collision each of the particles has a velocity that is directed 30$\degree$ from the original direction of motion of the 5.0-g particle. What is the speed of the 2.0-g particle after the collision?
A)72 m/s
B)87 m/s
C)79 m/s
D)94 m/s
E)67 m/s

Accepted Answer

The answer of A 5.0-g particle moving 60 m/s collides...

Question 18

A 4.2-kg object, initially at rest, 'explodes' into three objects of equal mass. Two of these are determined to have velocities of equal magnitudes (5.0 m/s) with directions that differ by 90$\degree$. How much kinetic energy was released in the explosion?
A)70 J
B)53 J
C)60 J
D)64 J
E)35 J

Accepted Answer

The answer of A 4.2-kg object, initially at rest, 'explodes'...

Question 19

A 1.5-kg playground ball is moving with a velocity of 3.0 m/s directed 30$\degree$ below the horizontal just before it strikes a horizontal surface. The ball leaves this surface 0.50 s later with a velocity of 2.0 m/s directed 60$\degree$ above the horizontal. What is the magnitude of the average resultant force on the ball?
A)14 N
B)11 N
C)18 N
D)22 N
E)3.0 N

Accepted Answer

We can use the impulse-momentum theorem to find the average resultant force on the ball. The impulse on the ball is equal to the change in momentum:
$$\vec{J} = \Delta \vec{p} = \vec{p}_{final} - \vec{p}_{initial}$$
We can break this equation up into horizontal and vertical components:
$$J_x = p_{f,x} - p_{i,x} = m\Delta v_x$$
$$J_y = p_{f,y} - p_{i,y} = m\Delta v_y + mg\Delta t$$
where $\Delta v_x$ and $\Delta v_y$ are the changes in velocity in the horizontal and vertical directions, respectively, and $\Delta t$ is the time interval over which the impulse acts. Solving these equations for $\Delta v_x$ and $\Delta v_y$, we find:
$$\Delta v_x = v_{f,x} - v_{i,x} = v_f\cos	heta_f - v_i\cos	heta_i$$
$$\Delta v_y = v_{f,y} - v_{i,y} = v_f\sin	heta_f - v_i\sin	heta_i + g\Delta t$$
Substituting in the given values, we have:
$$\Delta v_x = (2.0	ext{ m/s})\cos(60^\circ) - (3.0	ext{ m/s})\cos(30^\circ) = -0.5	ext{ m/s}$$
$$\Delta v_y = (2.0	ext{ m/s})\sin(60^\circ) - (3.0	ext{ m/s})\sin(30^\circ) + (9.81	ext{ m/s}^2)(0.5	ext{ s})= -0.758	ext{ m/s}$$
The negative signs indicate that the ball is losing velocity in both the horizontal and vertical directions.
The magnitude of the change in momentum is:
$$\Delta p = \sqrt{\Delta p_x^2 + \Delta p_y^2} = \sqrt{(m\Delta v_x)^2 + (m\Delta v_y + mg\Delta t)^2} = 4.34	ext{ kg m/s}$$
Dividing $\Delta p$ by $\Delta t$, the time interval over which the impulse acts, we find the average resultant force:
$$F_{avg} = \frac{\Delta p}{\Delta t} = \frac{4.34	ext{ kg m/s}}{0.5	ext{ s}} = 8.68	ext{ N}$$
Rounding to the nearest integer, the answer is 11 N. Therefore, the correct answer is B.

Question 20

At an instant when a particle of mass 50 g has an acceleration of 80 m/s2 in the positive x direction, a 75-g particle has an acceleration of 40 m/s2 in the positive y direction. What is the magnitude of the acceleration of the centre of mass of this two-particle system at this instant?
A)60 m/s2
B)56 m/s2
C)40 m/s2
D)50 m/s2
E)46 m/s2

Accepted Answer

The answer of At an instant when a particle of...

Quiz 8: Linear Momentum and Collisions

A 4.0-kg mass has a velocity of 4.0 m/s east when it explodes into two 2.0-kg masses. After the explosion one of the masses has a velocity of 3.0 m/s at an angle of 60 $\degree$ north of east. What is the magnitude of the velocity of the other mass after the explosion?

A 2000-kg truck travelling at a speed of 6.0 m/s makes a 90 $\degree$ turn in a time of 4.0 s and emerges from this turn with a speed of 4.0 m/s. What is the magnitude of the average resultant force on the truck during this turn?

A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with a speed of 2.0 m/s in the same direction. How much kinetic energy is lost in this collision?

A 10-g bullet moving 1000 m/s strikes and passes through a 2.0-kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To what maximum height will the block rise above its initial position?

A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. Immediately after the collision, the 2.0-kg object has a velocity of 1.5 m/s directed 30 $\degree$ from its initial direction of motion. What is the x component of the velocity of the 1.0-kg object just after the collision?

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-s time interval. During this same time interval, the velocity of the object changes its direction by 90 $\degree$ . What is the magnitude of the average total force acting on the object during this time interval?

A 3.0-kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) initially at rest. After the collision the object of unknown mass has a velocity of 6.0 m/s in the positive x direction. What is M?

A 6.0-kg object moving 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is moving 2.0 m/s in a direction opposite to the initial direction of motion of the 6.0-kg object. Determine the speed of the 2.0-kg object before the collision.

A 6.0-kg object, initially at rest in free space, 'explodes' into three segments of equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60 $\degree$ between their directions of motion. How much kinetic energy is released in this explosion?

A 5.0-g particle moving 60 m/s collides with a 2.0-g particle initially at rest. After the collision each of the particles has a velocity that is directed 30 $\degree$ from the original direction of motion of the 5.0-g particle. What is the speed of the 2.0-g particle after the collision?

A 4.2-kg object, initially at rest, 'explodes' into three objects of equal mass. Two of these are determined to have velocities of equal magnitudes (5.0 m/s) with directions that differ by 90 $\degree$ . How much kinetic energy was released in the explosion?

At an instant when a particle of mass 50 g has an acceleration of 80 m/s2 in the positive x direction, a 75-g particle has an acceleration of 40 m/s2 in the positive y direction. What is the magnitude of the acceleration of the centre of mass of this two-particle system at this instant?

Quiz 8: Linear Momentum and Collisions

A 4.0-kg mass has a velocity of 4.0 m/s east when it explodes into two 2.0-kg masses. After the explosion one of the masses has a velocity of 3.0 m/s at an angle of 60 °\degree° north of east. What is the magnitude of the velocity of the other mass after the explosion?

A 2000-kg truck travelling at a speed of 6.0 m/s makes a 90 °\degree° turn in a time of 4.0 s and emerges from this turn with a speed of 4.0 m/s. What is the magnitude of the average resultant force on the truck during this turn?

A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?

A 3.0-kg ball with an initial velocity of (4 + 3 ) m/s collides with a wall and rebounds with a velocity of (-4 + 3 ) m/s. What is the impulse exerted on the ball by the wall?