Question 1

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together.Assume that any other unbalanced forces are negligible.
(a)What is the speed of the wreckage just after the collision?
(b)In what direction does the wreckage move just after the collision?

Accepted Answer

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Question 2

A 480-kg car moving at 14.4 m/s hits from behind a 570-kg car moving at 13.3 m/s in the same direction.If the new speed of the heavier car is 14.0 m/s,what is the speed of the lighter car after the collision,assuming that any unbalanced forces on the system are negligibly small?

Accepted Answer

Momentum is conserved in the collision, so the total momentum before the collision is equal to the total momentum after the collision. The total momentum before the collision is:$$(480 	ext{ kg})(14.4 	ext{ m/s}) + (570 	ext{ kg})(13.3 	ext{ m/s}) = 14,016 	ext{ kg m/s}$$The total momentum after the collision is:$$(480 	ext{ kg})(14.0 	ext{ m/s}) + (570 	ext{ kg})(v) = 14,016 	ext{ kg m/s}$$where v is the speed of the lighter car after the collision. Solving for v, we get:$$v = \frac{14,016 	ext{ kg m/s} - (480 	ext{ kg})(14.0 	ext{ m/s})}{570 	ext{ kg}} = 13.6 	ext{ m/s}$$Therefore, the speed of the lighter car after the collision is 13.6 m/s.

Question 3

A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table.The initially moving billiard ball deflects 30.0° from its original direction.What is the speed of the initially stationary billiard ball after the collision?

Accepted Answer

In a perfectly elastic collision between two identical masses, the initially stationary ball will move at a speed equal to the initial speed of the moving ball times the sine of the deflection angle. Thus, the speed is 3.00 m/s * sin(30°) = 1.50 m/s.

Question 4

A pool player is attempting a fancy shot.He hits the cue ball giving it a speed of 5.57 m/s and directs its center on a path tangent to the surface of the target ball having the same mass as the cue ball.After the collision (on a frictionless table)the initially-stationary ball moves with a speed of 4.82 m/s.After the collision,the new speed of the cue ball and the relative direction of the balls are closest to

Accepted Answer

The conservation of momentum and kinetic energy in elastic collisions applies here. Since the balls have the same mass and the table is frictionless, the velocities can be calculated using these principles. The speed of the cue ball after the collision can be found using the conservation of momentum and the fact that the collision is elastic. The direction at 90° indicates that the collision was perfectly elastic and the balls had equal masses, which is a characteristic outcome for such collisions when one ball strikes another at a tangent.

Question 5

A 620-g object traveling at 2.1 m/s collides head-on with a 320-g object traveling in the opposite direction at 3.8 m/s.If the collision is perfectly elastic,what is the change in the kinetic energy of the 620-g object?

Accepted Answer

Explanation:The correct answer is A. The change in kinetic energy of the 620-g object is -0.23 J. This is because the collision is perfectly elastic, which means that the total kinetic energy of the system is conserved. The initial kinetic energy of the system is:$$KE_i = (1/2)mv_1^2 + (1/2)mv_2^2 = (1/2)(0.62 kg)(2.1 m/s)^2 + (1/2)(0.32 kg)(3.8 m/s)^2 = 2.26 J$$The final kinetic energy of the system is:$$KE_f = (1/2)mv_3^2 + (1/2)mv_4^2 = (1/2)(0.62 kg)(v_3)^2 + (1/2)(0.32 kg)(v_4)^2$$where v_3 and v_4 are the velocities of the 620-g and 320-g objects after the collision, respectively.Using the conservation of momentum, we can find that:$$v_3 = (m_1v_1 - m_2v_2)/(m_1 + m_2) = (0.62 kg)(2.1 m/s) - (0.32 kg)(3.8 m/s))/(0.62 kg + 0.32 kg) = 0.98 m/s$$$$v_4 = (m_2v_2 - m_1v_1)/(m_1 + m_2) = (0.32 kg)(3.8 m/s) - (0.62 kg)(2.1 m/s))/(0.62 kg + 0.32 kg) = -2.82 m/s$$Therefore, the change in kinetic energy of the 620-g object is:$$KE_f - KE_i = (1/2)(0.62 kg)(0.98 m/s)^2 + (1/2)(0.32 kg)(-2.82 m/s)^2 - 2.26 J = -0.23 J$$

Question 6

A 2.3-kg object traveling at 6.1 m/s collides head-on with a 3.5-kg object traveling in the opposite direction at 4.8 m/s.If the collision is perfectly elastic,what is the final speed of the  ?

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Question 7

In the figure,determine the character of the collision.The masses of the blocks,and the velocities before and after are given.The collision is

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The answer of In the figure,determine the character of the...

Question 8

A car of mass 1689 kg collides head-on with a parked truck of mass 2000 kg.Spring mounted bumpers ensure that the collision is essentially elastic.If the velocity of the truck is  (in the same direction as the car's initial velocity)after the collision,what was the initial speed of the car?

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Question 9

A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally to the right at 12.0 m/s.The stone rebounds to the left with a speed of 8.50 m/s,and the ball swings to a maximum height h above its original level.The value of h is closest to

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Question 10

Two objects of the same mass move along the same line in opposite directions.The first mass is moving with speed v.The objects collide,stick together,and move with speed 0.100v in the direction of the velocity of the first mass before the collision.What was the speed of the second mass before the collision?

Accepted Answer

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Question 11

A block of mass m = 8.40 kg,moving on a horizontal frictionless surface with a speed 4.20 m/s,makes a perfectly elastic collision with a block of mass M at rest.After the collision,the 8.40 block recoils with a speed of 0.400 m/s.In the figure,the blocks are in contact for 0.200 s.The magnitude of the average force on the 8.40-kg block,while the two blocks are in contact,is closest to

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Question 12

A 15-g bullet is shot vertically into an 2-kg block.The block lifts upward 8.0 mm (see the figure).The bullet penetrates the block and comes to rest in it in a time interval of 0.0010 s.Assume the force on the bullet is constant during penetration and that air resistance is negligible.The initial kinetic energy of the bullet is closest to

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Question 13

A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.After the collision,the 2.00-kg object has a velocity 5.00 m/s to the west.How much kinetic energy was lost during the collision?

Accepted Answer

The kinetic energy before the collision can be calculated using the formula $KE = \frac{1}{2}mv^2$ for each object and then summing them up. For the 2.00-kg object, $KE_1 = \frac{1}{2}(2.00\, \text{kg})(20.0\, \text{m/s})^2 = 400\, \text{J}$. For the 3.00-kg object, $KE_2 = \frac{1}{2}(3.00\, \text{kg})(10.0\, \text{m/s})^2 = 150\, \text{J}$. The total initial kinetic energy is $400\, \text{J} + 150\, \text{J} = 550\, \text{J}$.After the collision, the 2.00-kg object has a velocity of 5.00 m/s to the west. The kinetic energy of this object is $KE_{1, \text{after}} = \frac{1}{2}(2.00\, \text{kg})(5.00\, \text{m/s})^2 = 25\, \text{J}$. To find the velocity of the 3.00-kg object after the collision, we use the conservation of momentum, since momentum is conserved in collisions. The initial momentum is $p_{\text{initial}} = (2.00\, \text{kg})(20.0\, \text{m/s}) + (3.00\, \text{kg})(-10.0\, \text{m/s}) = 40\, \text{kg}\cdot\text{m/s} - 30\, \text{kg}\cdot\text{m/s} = 10\, \text{kg}\cdot\text{m/s}$.After the collision, the momentum of the 2.00-kg object is $(2.00\, \text{kg})(-5.00\, \text{m/s}) = -10\, \text{kg}\cdot\text{m/s}$. To conserve momentum, the 3.00-kg object must have a momentum of $10\, \text{kg}\cdot\text{m/s} + 10\, \text{kg}\cdot\text{m/s} = 20\, \text{kg}\cdot\text{m/s}$, giving it a velocity of $20\, \text{kg}\cdot\text{m/s} / 3.00\, \text{kg} = 6.67\, \text{m/s}$. The kinetic energy of the 3.00-kg object after the collision is $KE_{2, \text{after}} = \frac{1}{2}(3.00\, \text{kg})(6.67\, \text{m/s})^2 \approx 66.7\, \text{J}$.The total kinetic energy after the collision is $25\, \text{J} + 66.7\, \text{J} = 91.7\, \text{J}$. The kinetic energy lost during the collision is $550\, \text{J} - 91.7\, \text{J} = 458.3\, \text{J}$, which rounds to $458\, \text{J}$, making option B correct.

Question 14

An 8.0-g bullet is shot into a 4.0-kg block,at rest on a frictionless horizontal surface (see the figure).The bullet remains lodged in the block.The block moves into an ideal massless spring and compresses it by 8.7 cm.The spring constant of the spring is 2400 N/m.The initial velocity of the bullet is closest to

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Question 15

In the figure,determine the character of the collision.The masses of the blocks,and the velocities before and after are given,and no other unbalanced forces act on these blocks.The collision is

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The answer of In the figure,determine the character of the...

Question 16

A 1000-kg car approaches an intersection traveling north at 20.0 m/s.A 1200-kg car approaches the same intersection traveling east at 22.0 m/s.The two cars collide at the intersection and lock together.Ignoring any external forces that act on the cars during the collision,what is the velocity of the cars immediately after the collision?

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Question 17

On a frictionless horizontal table,two blocks (A of mass 2.00 kg and B of mass 3.00 kg)are pressed together against an ideal massless spring that stores 75.0 J of elastic potential energy.The blocks are not attached to the spring and are free to move free of it once they are released from rest.The maximum speed achieved by each block is closest to

Accepted Answer

The maximum speed of each block can be found using the conservation of energy principle. The total elastic potential energy stored in the spring is converted into the kinetic energy of the two blocks. The kinetic energy ($KE$) is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity of the block. Since the total energy is conserved, the sum of the kinetic energies of both blocks equals the initial potential energy stored in the spring, i.e., $75.0 J$.For two blocks of different masses, the block with the smaller mass (A, 2.00 kg) will achieve a higher speed than the block with the larger mass (B, 3.00 kg) because the same amount of energy results in a higher velocity for the lighter object. The correct speeds, as calculated using the conservation of energy and taking into account the mass of each block, are 6.71 m/s for block A and 4.47 m/s for block B, making option A the correct choice.

Question 18

Two automobiles traveling at right angles to each other collide and stick together.Car A has a mass of 1200 kg and had a speed of 25 m/s before the collision.Car B has a mass of 1600 kg.The skid marks show that,immediately after the collision,the wreckage was moving in a direction making an angle of 40° with the original direction of car A.What was the speed of car B before the collision,assuming that any other unbalanced forces are negligible?

Accepted Answer

Explanation:Momentum is conserved in both the x and y directions.In the x direction:(1200 kg)(25 m/s) + (1600 kg)(v_B)cos40° = (2800 kg)(v_f)cos40°In the y direction:(1600 kg)(v_B)sin40° = (2800 kg)(v_f)sin40°Dividing the first equation by the second equation, we get:(1200 kg)(25 m/s) + (1600 kg)(v_B)cos40° = (2800 kg)(v_f)cos40°(1200 kg)(25 m/s) + (1600 kg)(v_B)cos40° = (2800 kg)(v_f)cos40°v_B = 16 m/s

Question 19

A car heading north collides at an intersection with a truck of the same mass as the car heading east.If they lock together and travel at 28 m/s at 46° north of east just after the collision,how fast was the car initially traveling? Assume that any other unbalanced forces are negligible.

Accepted Answer

The conservation of momentum principle is used here. The momentum before the collision must equal the momentum after the collision. Let $v_c$ be the velocity of the car, and $v_t = 28\,m/s$ (the combined velocity after the collision). The angle given is 46° north of east, which means we can decompose the final velocity into north and east components using trigonometry. The east component (truck's direction) is $v_t \cos(46°)$ and the north component (car's direction) is $v_t \sin(46°)$. Since the masses are the same and cancel out, we can set the initial velocity of the car equal to its component of the final velocity: $v_c = v_t \sin(46°)$. Plugging in the numbers: $v_c = 28 \sin(46°) \approx 20\,m/s$ for the north direction. However, this calculation mistake led to the wrong option being chosen. Correcting the approach: The correct method involves using the conservation of momentum in vector form, where the momentum of the car initially (in the north direction) plus the momentum of the truck initially (in the east direction) equals the combined momentum after the collision. The error in calculation led to an incorrect conclusion. The correct calculation should consider both the north and east components of momentum and solve for the initial speed of the car using the given final speed and angle. Given the mistake in the initial explanation, the correct approach involves solving for the car's initial speed using the conservation of momentum and the given final velocity and direction of the combined masses. Without the correct calculations provided, the initial choice was incorrect. The correct answer, based on the conservation of momentum and the information provided, needs to be recalculated correctly.

Question 20

Two ice skaters push off against one another starting from a stationary position.The 45.0-kg skater acquires a speed of 0.375 m/s.What speed does the 60.0-kg skater acquire? Assume that any other unbalanced forces during the collision are negligible.

Accepted Answer

By conservation of momentum in the x-direction, we have:

(m1)(v1) + (m2)(v2) = 0

where m1 and v1 are the mass and velocity of the 45.0-kg skater, and m2 and v2 are the mass and velocity of the 60.0-kg skater. Initially, both skaters are at rest, so v1 = 0. After the collision, the 45.0-kg skater has a speed of 0.375 m/s. Substituting these values into the equation and solving for v2 gives:

(45.0 kg)(0.375 m/s) + (60.0 kg)(v2) = 0

16.875 + 60v2 = 0

v2 = -0.28125 m/s

The negative sign means that the 60.0-kg skater is moving in the opposite direction to the 45.0-kg skater. Taking the absolute value gives:

|v2| = 0.281 m/s

Therefore, the correct answer is B.

Quiz 8: Momentum,impulse,and Collisions