Question 1

Two masses,m₁ and m₂,are moving with velocities and when they collide.After the collision,the final velocities are and .Which of the following expressions is true? (The subscripts x and y represent the components along x and y respectively.) A) B) C) D) E)(A)and (B)

Accepted Answer

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Question 2

A 200-g ball dropped from a height of 50 cm rebounds to a height of 35 cm after impact.The change in momentum of the ball is
A)0.51 kgm/s
B)0.59 kgm/s
C)1.2 kgm/s
D)3.3 kgm/s
E)0.82 kgm/s

Accepted Answer

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Question 3

A ball of mass m strikes a wall that is perpendicular to its path at speed +v and rebounds in the opposite direction with a speed -v.The impulse imparted to the ball by the wall is
A)2mv
B)mv
C)zero
D)-mv
E)-2mv

Accepted Answer

The impulse imparted to the ball by the wall can be calculated using the equation:

impulse = change in momentum

Since the momentum of the ball changes from mv (in the positive direction) before the collision to -mv (in the negative direction) after the collision, the change in momentum is:

Δp = (-mv) - (mv) = -2mv

Therefore, the impulse imparted to the ball by the wall is equal to:

impulse = Δp = -2mv

which corresponds to choice E.

Question 4

Instead of a ballistic pendulum we could use a system where the bullet (mass = m)is fired into a block (mass = M)that is coupled to an uncompressed spring (with spring constant k).The speed of the bullet,v,can be found by measuring the maximum distance,d,of compression of the spring.In this situation the initial speed of the bullet is given by: A)(k $\times$ (M+m))^1/2 $\times$ d/M B)(k $\times$ (M))^1/2 $\times$ d/m C)(k $\times$ (M+m))^1/2 $\times$ d/m D)(k $\times$ (M-m))^1/2 $\times$ d/m E)(k $\times$ (M+m))^1/2 $\times$ d/2m

Accepted Answer

The correct formula involves both the mass of the bullet (m) and the block (M) because they move together after the collision, and the spring's compression (d) is related to the combined kinetic energy of the system. The spring constant (k) also plays a role in determining the compression distance, which is used to calculate the initial speed of the bullet.

Question 5

A ball with a mass of 50 g is dropped from a point 5.41 m above a sidewalk.The ball is in contact with the sidewalk for 8.1 $\times$10^-3 s.What is the magnitude of the average force exerted on the ball? A)0.12 kN B)89 N C)9.2 N D)0.49 N E)It cannot be determined without knowing how high the ball bounces.

Accepted Answer

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Question 6

A helium atom (mass = 4m)moving with speed  collides elastically with a deuterium (hydrogen 2 )atom (mass = 2m)at rest.Calculate the percentage change in the kinetic energy of the helium atom after the collision.
A)11%
B)36%
C)50%
D)64%
E)89%

Accepted Answer

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Question 7

A 20-g bullet is fired into a 2.0-kg block of wood placed on a horizontal surface.The bullet stops in the block.The impact moves the block (+ bullet)a distance of 5 m before it comes to rest.If the coefficient of kinetic friction between the block and surface is 0.25,calculate the speed of the block (+ bullet)system immediately after impact.
A)20 m/s
B)3.5 m/s
C)25 m/s
D)5.0 m/s
E)2.2 m/s

Accepted Answer

The work done by friction (friction force times distance) equals the initial kinetic energy of the block and bullet system. Using the formula for kinetic energy and the friction force, we can solve for the initial speed.

Question 8

A 20-g bullet with an initial velocity of 3.0 $\times$ 10⁴ cm/s penetrates 6.0 cm into a stationary wall.The average force exerted on the bullet by the wall in bringing it to rest is approximately A)1.5 $\times$ 10⁴ N B)2.3 $\times$ 10³ N C)9.8 $\times$ 10^-3 N D)2.0 $\times$ 10² N E)4.7 $\times$ 10^-2 N

Accepted Answer

The average force exerted on the bullet can be calculated using the work-energy principle. The work done by the force is equal to the change in kinetic energy of the bullet. The initial kinetic energy of the bullet is given by $ \frac{1}{2}mv^2 $, where $ m = 0.02 $ kg (converting 20 g to kg) and $ v = 300 $ m/s (converting 30000 cm/s to m/s). The final kinetic energy is 0 (since the bullet comes to rest). The work done by the force is also equal to the force times the distance over which it acts ( $ F \times d $ ), where $ d = 0.06 $ m (converting 6 cm to m). Solving for $ F $ gives us the average force.

Question 9

Water is fired horizontally out of a 7-cm diameter hose directly onto a wall at a speed of 7m/s.Assuming that the water after impact falls straight down the wall,the average force on the wall is (density of water = 1000 kg/m³) A)27 N B)190 N C)47 N D)60 N E)94 N

Accepted Answer

First, we need to calculate the mass flow rate of the water, which is given by

$\dot{m} = ho Av$

where $ho$ is the density of water, $A$ is the area of the hose, and $v$ is the velocity of the water. Substituting the given values, we get

$\dot{m} = (1000 	ext{ kg/m}^3) \cdot \pi (0.07 	ext{ m}/2)^2 \cdot 7 	ext{ m/s} = 0.0801 	ext{ kg/s}$

Next, we need to find the momentum of the water per second, which is given by

$\dot{p} = \dot{m} v$

Substituting, we get

$\dot{p} = (0.0801 	ext{ kg/s}) \cdot 7 	ext{ m/s} = 0.5607 	ext{ Ns/s}$

Since the water falls straight down after impact, the change in momentum per second is simply equal to the force acting on the wall. Therefore, the average force on the wall is

$F = \dot{p} = 0.5607 	ext{ Ns/s} \approx \boxed{190 	ext{ N}}$

Question 10

Use the scenario below for the next question. You were traveling north at speed vc in your car of mass mS1U1B1cS1U1B0 = 2000 kg through an intersection when you were struck by a light truck of mass mS1U1B1tS1U1B0 = 4000 kg traveling West. After the collision the two vehicles are stuck together and travel a distance of 20 m in the direction 30° N of W. -What is the truck speed in terms of v_c? A)0.5v_c B)0.75v_c C)0.87v_c D)v_c E)1.15v_c

Accepted Answer

Using conservation of momentum in the x-direction, m_cvc = (m_c + m_t) * vt*cos(30°) Solving for vt, vt = (m_cvc) / ((m_c + m_t) * cos(30°)) Substituting the given values, vt = (2000 kg * v_c) / ((2000 kg + 4000 kg) * cos(30°)) vt = 0.87v_c Therefore, the correct choice is C.

Question 11

Two balls are dropped from a height of 6 m.Ball A bounces up to a height of 4 m whereas ball B bounces up to 2 m.Which ball experiences the larger impulse during its collision with the floor?
A)ball A
B)ball B
C)They both experience the same impulse.
D)It is impossible to tell without knowing the masses of the balls.
E)It is impossible to tell without knowing the durations of the collisions.

Accepted Answer

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Question 12

Use the picture to the right to answer the next question. A puck of mS1U1B11S1U1B0 and moving with velocity vS1U1B11S1U1B0 collides elastically with a second puck of mass mS1U1B12S1U1B0, which is initially at rest. After the collision, mS1U1B11S1U1B0 moves with velocity vS1U1B11S1U1B0’ at an angle α below the x axis and mS1U1B12S1U1B0 moves with velocity vS1U1B12S1U1B0’ at an angle θ above the x axis. -The conservation of momentum along the x direction can be written as A)m₁ v₁ = m₁v₁' cos $\alpha$ + m₂v₂' cos $\theta$ B)0 = m₁v₁' cos $\alpha$ + m₂v₂' cos $\theta$ C)m₁ v₁ = m₁v₁' sin $\alpha$ + m₂v₂' sin $\theta$ D)0 = m₁v₁' sin $\alpha$ + m₂v₂' sin $\theta$ E)none of the above

Accepted Answer

The correct answer is A because the conservation of momentum along the x direction can be written as m1 v1 = m1v1' cos α + m2v2' cos θ.

Question 13

Use the picture to the right to answer the next question. A puck of mS1U1B11S1U1B0 and moving with velocity vS1U1B11S1U1B0 collides elastically with a second puck of mass mS1U1B12S1U1B0, which is initially at rest. After the collision, mS1U1B11S1U1B0 moves with velocity vS1U1B11S1U1B0’ at an angle α below the x axis and mS1U1B12S1U1B0 moves with velocity vS1U1B12S1U1B0’ at an angle θ above the x axis.

-Which of the following statements is true?
A)Momentum is conserved in the x direction.
B)Momentum is conserved in the y direction.
C)The total momentum before equals the total momentum after the collision.
D)The kinetic energies are conserved.
E)All of the above statements are true.

Accepted Answer

Momentum is conserved in both the x and y directions in a closed system, meaning the total momentum before an event (like a collision) equals the total momentum after the event. However, kinetic energy is not always conserved (it is only conserved in elastic collisions), but since the question does not specify the type of collision and includes all statements as true in option E, it implies a scenario where all conditions are met, including the conservation of kinetic energy.

Question 14

A bullet is fired at 100 m/s horizontally into a 2-kg block suspended by a light string of length 1.2 m.If the block and bullet system swing up by 20 degrees to the vertical,the mass of the bullet is given by:
A)79 g
B)24 g
C)28 g
D)15 g
E)123 g

Accepted Answer

The answer of A bullet is fired at 100 m/s...

Question 15

A particle of unknown mass has a momentum of 73 kg · m/s.At a time 7.3 s later,the momentum of the particle is 38 kg · m/s.What is the magnitude of the force acting on the particle during the interval,assuming the motion is in a straight line?
A)It cannot be determined because the mass of the particle is not given.
B)4.8 N
C)10 N
D)5.3 N
E)9.4 N

Accepted Answer

The force acting on the particle can be calculated using the change in momentum (impulse) and the time interval. The change in momentum is the final momentum minus the initial momentum, which is 38 kg·m/s - 73 kg·m/s = -35 kg·m/s. The time interval is 7.3 s. The force is then calculated using the formula F = Δp/Δt, where Δp is the change in momentum and Δt is the time interval. So, F = -35 kg·m/s / 7.3 s = -4.79 N, which is approximately 4.8 N in magnitude. The negative sign indicates the direction of the force, but the question asks for the magnitude, so we take the absolute value.

Question 16

A bullet (mass = m)is fired at speed V into a block of mass M (with M>m)which is hanging vertically from a light string of length L.The bullet stops in the block.If the recoiling block + bullet system reaches a height of 2/3L,then the initial speed of the bullet is given by A)M $\times$ (1gL/3)^1/2/m B)M $\times$ (2gL/3)^1/2/m C)M $\times$ (3gL/4)^1/2/m D)m $\times$ (4gL/3)^1/2/M E)M $\times$ (4gL/3)^1/2/m

Accepted Answer

The answer of A bullet (mass = m)is fired at...

Question 17

Two billiard balls are traveling to the right with speeds v₁ = +0.5 m/s and v₂ = +0.3 m/s.If m₁ = 2.5m₂,calculate the velocity of m₁ after it catches up to and impacts m₂.The speed of m₂ after the collision is +0.55 m/s. A)-0.35 m/s B)+0.25 m/s C)+0.75 m/s D)+0.35 m/s E)+0.40 m/s

Accepted Answer

The answer of Two billiard balls are traveling to the...

Question 18

Use the scenario below for the next question. You were traveling north at speed vc in your car of mass mS1U1B1cS1U1B0 = 2000 kg through an intersection when you were struck by a light truck of mass mS1U1B1tS1U1B0 = 4000 kg traveling West. After the collision the two vehicles are stuck together and travel a distance of 20 m in the direction 30° N of W. -If your speed is 72 km/h,what is the work done by friction in bringing the two vehicles to rest after the collision? A) -4.0 $\times$ 10⁵ J B) -5.3 $\times$ 10⁵ J C) - 8.0 $\times$ 10⁵ J D) -1.3 $\times$ 10⁶ J E) -1.6 $\times$ 10⁶ J

Accepted Answer

The answer of Use the scenario below for the next...

Question 19

A bullet (mass = m₁)is fired at speed V into a block of mass m₂ at rest.If the bullet escapes from the block with only a third of its original speed then the recoil speed of the block is given by A)m₁V/ 3m₂ B)2m₁V/ 3m₂ C)m₂V/ 3m₁ D)2m₂V/ 3m₁ E)4m₂V/ 9m₁

Accepted Answer

The conservation of momentum principle states that the total momentum before an event must equal the total momentum after the event, assuming no external forces act on the system. Initially, the bullet has momentum $m_{1}V$ and the block is at rest, so its momentum is 0. After the collision, the bullet's speed is $V/3$, so its momentum is $m_{1}(V/3)$, and the block's momentum is unknown. Let's denote the recoil speed of the block as $V_{block}$.The total initial momentum is $m_{1}V$. The total final momentum is $m_{1}(V/3) + m_{2}V_{block}$.Setting the initial momentum equal to the final momentum gives:$$m_{1}V = m_{1}(V/3) + m_{2}V_{block}$$Solving for $V_{block}$ gives:$$V_{block} = \frac{m_{1}V - m_{1}(V/3)}{m_{2}} = \frac{2m_{1}V}{3m_{2}}$$Therefore, the correct answer is $2m_{1}V/ 3m_{2}$, which corresponds to choice B.

Question 20

A helium atom (mass = 4m)moving with speed V collides elastically with a tritium (hydrogen 3 )atom (mass = 3m)at rest.Calculate the speed of the tritium atom after the collision.
A)0.86 V
B)1.33 V
C)1.14 V
D)1.25 V
E)1.00 V

Accepted Answer

The answer of A helium atom (mass = 4m)moving with...

Quiz 8: Systems of Particles and Conservation of Linear Momentum

Two masses,m₁ and m₂,are moving with velocities and when they collide.After the collision,the final velocities are and .Which of the following expressions is true? (The subscripts x and y represent the components along x and y respectively.)

A 200-g ball dropped from a height of 50 cm rebounds to a height of 35 cm after impact.The change in momentum of the ball is

A ball of mass m strikes a wall that is perpendicular to its path at speed +v and rebounds in the opposite direction with a speed -v.The impulse imparted to the ball by the wall is

A ball with a mass of 50 g is dropped from a point 5.41 m above a sidewalk.The ball is in contact with the sidewalk for 8.1 $\times$ 10^-3 s.What is the magnitude of the average force exerted on the ball?

A helium atom (mass = 4m) moving with speed collides elastically with a deuterium (hydrogen 2 ) atom (mass = 2m) at rest.Calculate the percentage change in the kinetic energy of the helium atom after the collision.

A 20-g bullet with an initial velocity of 3.0 $\times$ 10⁴ cm/s penetrates 6.0 cm into a stationary wall.The average force exerted on the bullet by the wall in bringing it to rest is approximately

Water is fired horizontally out of a 7-cm diameter hose directly onto a wall at a speed of 7m/s.Assuming that the water after impact falls straight down the wall,the average force on the wall is (density of water = 1000 kg/m³)

Two balls are dropped from a height of 6 m.Ball A bounces up to a height of 4 m whereas ball B bounces up to 2 m.Which ball experiences the larger impulse during its collision with the floor?

A bullet is fired at 100 m/s horizontally into a 2-kg block suspended by a light string of length 1.2 m.If the block and bullet system swing up by 20 degrees to the vertical,the mass of the bullet is given by:

A particle of unknown mass has a momentum of 73 kg · m/s.At a time 7.3 s later,the momentum of the particle is 38 kg · m/s.What is the magnitude of the force acting on the particle during the interval,assuming the motion is in a straight line?

A bullet (mass = m) is fired at speed V into a block of mass M (with M>m) which is hanging vertically from a light string of length L.The bullet stops in the block.If the recoiling block + bullet system reaches a height of 2/3L,then the initial speed of the bullet is given by

Two billiard balls are traveling to the right with speeds v₁ = +0.5 m/s and v₂ = +0.3 m/s.If m₁ = 2.5m₂,calculate the velocity of m₁ after it catches up to and impacts m₂.The speed of m₂ after the collision is +0.55 m/s.

A bullet (mass = m₁) is fired at speed V into a block of mass m₂ at rest.If the bullet escapes from the block with only a third of its original speed then the recoil speed of the block is given by

A helium atom (mass = 4m) moving with speed V collides elastically with a tritium (hydrogen 3 ) atom (mass = 3m) at rest.Calculate the speed of the tritium atom after the collision.