Question 1

A car (of mass 1500 kg)and a small truck (of mass 2000 kg)collide suddenly at right angles at an icy intersection.The car was originally traveling eastward at 20.0 m/s and the truck was traveling northward at 20.0 m/s just before the collision took place.What is the speed of the combined wreck just after the collision if the vehicles become enmeshed?

Accepted Answer

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Question 2

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.In what direction does the wreckage move just after the collision?

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 3

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.What is the speed of the wreckage just after the collision?

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 4

An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east.Immediately after the collision,the man and boy are seen to be moving together at 2.5 m/s in a direction 60° north of east.How fast was the boy moving just before the collision?

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Question 5

A puck with a mass m₁ = 50 g moving at 1.0 m/s approaches a stationary puck with a mass m₂ = 100 g on a frictionless air table,and they undergo a sudden two-dimensional elastic collision.After the collision,both pucks have identical speeds,but travel in different directions.What is the angle between the original and final paths of puck with mass m₁?

Accepted Answer

Since both pucks have identical speeds after the collision, their paths must make an angle of 90° with each other. This is because in an elastic collision, the momentum and kinetic energy are conserved, which means that the final kinetic energy is the same as the initial kinetic energy, but the direction of the momentum can change. Therefore, the angle between the original and final paths is 90°.

Question 6

A batter hits a foul ball.The 0.14-kg baseball that was approaching him at 40 m/s leaves the bat at 30 m/s in a direction perpendicular to the line between the batter and the pitcher.What is the magnitude of the impulse exerted on the baseball?

Accepted Answer

Impulse is the change in momentum. The initial momentum is 0.14 kg * 40 m/s = 5.6 kg∙m/s. The final momentum is 0.14 kg * 30 m/s = 4.2 kg∙m/s. The impulse is the vector sum of these, which is √(5.6² + 4.2²) = 7.0 N∙s.

Question 7

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south.The two cars stick together after the collision.What is the speed of the cars after collision?

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Question 8

A fire hose is turned on the door of a burning building in order to knock the door down.This requires a force of 1000 N.If the hose delivers 40 kg of water per second,what is the minimum velocity of the stream needed to knock down the door,assuming the water doesn't bounce back?

Accepted Answer

We can use the equation F = m*v, where F is the force, m is the mass, and v is the velocity. Rearranging the equation to solve for v, we get v = F/m.
Plugging in the values, we get v = 1000 N / (40 kg/s) = 25 m/s. Therefore, the minimum velocity needed to knock down the door is 25 m/s, which corresponds to choice C.

Question 9

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s.It suddenly collides directly with a stationary seal of mass 200 kg.The seal grabs onto the whale and holds fast.What is the speed of these two sea creatures just after their collision? You can neglect any drag effects of the water during the collision.

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Question 10

A 900-kg car traveling 30.0° south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s.The cars stick together after colliding.What is the speed of the wreckage just after the collision?

Accepted Answer

To find the speed of the wreckage just after the collision, we use the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision. We break down the velocities into their components and calculate the final velocity vector of the combined mass.For the 900-kg car:- The eastward component of velocity: $12.0 \cos(30.0°) = 10.39 \, \text{m/s}$- The southward component of velocity: $12.0 \sin(30.0°) = 6.0 \, \text{m/s}$For the 750-kg car traveling north (which can be considered as having a negative southward velocity):- The northward velocity (or negative southward velocity): $-17.0 \, \text{m/s}$The total momentum in the eastward direction before the collision is:$900 \times 10.39 = 9351 \, \text{kg·m/s}$The total momentum in the southward direction before the collision is:$900 \times 6.0 + 750 \times (-17.0) = 5400 - 12750 = -7350 \, \text{kg·m/s}$The total mass of the wreckage is:$900 + 750 = 1650 \, \text{kg}$To find the velocity of the wreckage, we calculate the magnitude of the velocity vector using the eastward and southward components of momentum:$$v = \sqrt{\left(\frac{9351}{1650}\right)^2 + \left(\frac{-7350}{1650}\right)^2}$$$$v = \sqrt{(5.67)^2 + (-4.45)^2}$$$$v = \sqrt{32.17 + 19.80}$$$$v = \sqrt{51.97} \approx 7.21 \, \text{m/s}$$Therefore, the speed of the wreckage just after the collision is approximately 7.21 m/s.

Question 11

A 2200-kg auto moving northward at 12.0 m/s runs into a 3800-kg truck which is also moving northward,but at 5.00 m/s.If the vehicles lock bumpers,how fast are they moving just after the collision?

Accepted Answer

The answer of A 2200-kg auto moving northward at 12.0...

Question 12

A force of 5.3 N is needed to hold on to an umbrella in a strong wind.If the air molecules each have a mass of 4.7 ×10^-26 kg,and each one strikes the umbrella (without rebounding)with a speed of 2.0 m/s in the same direction,how many molecules strike the umbrella each second? Assume that the wind blows horizontally so that the gravitational force exerted on the air molecules can be neglected.

Accepted Answer

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Question 13

A 0.140-kg baseball is dropped from rest.It has a speed of 1.20 m/s just before it hits the ground,and it rebounds with an upward speed of 1.00 m/s.The ball is in contact with the ground for 0.0140 s.What is the average force exerted by the ground on the ball during the time of contact?

Accepted Answer

We can use the impulse-momentum theorem to solve this problem. The impulse-momentum theorem states that the impulse (change in momentum) of an object is equal to the net force acting on it multiplied by the time over which the force is applied. In equation form:

J = Δp = FΔt

where:
J = impulse
Δp = change in momentum
F = net force
Δt = time over which the force is applied

First, let's find the initial momentum of the ball:

p1 = m1v1 = (0.140 kg)(-1.20 m/s) = -0.168 kg*m/s

The negative sign indicates that the momentum is downward (the direction of gravity).

Next, let's find the final momentum of the ball:

p2 = m2v2 = (0.140 kg)(1.00 m/s) = 0.140 kg*m/s

The positive sign indicates that the momentum is upward (the direction of the rebound).

The change in momentum is then:

Δp = p2 - p1 = 0.140 kg*m/s - (-0.168 kg*m/s) = 0.308 kg*m/s

Finally, we can use the impulse-momentum theorem to solve for the average force:

F = Δp/Δt = 0.308 kg*m/s / 0.0140 s = 22.0 N upwards

Therefore, the correct answer is (C).

Question 14

Two vehicles approach a right angle intersection and then suddenly collide.After the collision,they become entangled.If their mass ratios were 1:4 and their respective speeds as they approached the intersection were both 13 m/s,find the magnitude and direction of the final velocity of the wreck.

Accepted Answer

To solve this problem, we use the principle of conservation of momentum. Since the vehicles become entangled, we treat them as a single system post-collision. The total momentum before the collision equals the total momentum after the collision.Let's denote the mass of the lighter vehicle as $m$ and the heavier vehicle as $4m$, given their mass ratio of 1:4. Both vehicles have a speed of 13 m/s before the collision. The momentum of each vehicle before the collision is the product of its mass and velocity. The total momentum in the system can be broken down into its components along the x and y axes (assuming one vehicle moves along the x-axis and the other along the y-axis).The momentum along the x-axis before the collision is $m \times 13$, and along the y-axis, it's $4m \times 13$.After the collision, the combined mass is $5m$, and let $v$ be the final velocity of the entangled wreck. The total momentum is conserved, so the magnitude of the total momentum before the collision equals the magnitude of the total momentum after the collision.The magnitude of the total momentum before the collision can be found using Pythagoras' theorem, considering the momentum components as perpendicular vectors:$$ \sqrt{(m \times 13)^2 + (4m \times 13)^2} = 13m \sqrt{1^2 + 4^2} = 13m \sqrt{17} $$Since the total momentum is conserved, the magnitude of the momentum after the collision is also $13m \sqrt{17}$. The final velocity $v$ of the entangled wreck is found by dividing the total momentum by the total mass ($5m$):$$ v = \frac{13m \sqrt{17}}{5m} = \frac{13 \sqrt{17}}{5} $$To find the angle, we note that the tangent of the angle ($\theta$) with respect to the original direction of the lighter car (x-axis) is given by the ratio of the y-component to the x-component of the momentum:$$ \tan(\theta) = \frac{4m \times 13}{m \times 13} = 4 $$Thus, $\theta = \arctan(4)$, which is approximately 76°.The magnitude of the final velocity can be calculated as:$$ v = \frac{13 \sqrt{17}}{5} \approx 11 \, \text{m/s} $$Therefore, the correct answer is 11 m/s at 76° with respect to the original direction of the lighter car.

Question 15

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s.It suddenly collides directly with a stationary seal of mass 200 kg.The seal grabs onto the whale and holds fast.What is the momentum of these two sea creatures just after their collision? You can neglect any drag effects of the water during the collision.

Accepted Answer

The answer of A 1000-kg whale swims horizontally to the...

Question 16

A 1200-kg car moving at $15.6 \mathrm {~m} / \mathrm { s }$ suddenly collides with a stationary car of mass $1500 \mathrm {~kg} \text {. }$ If the two vehicles lock together,what is their combined velocity immediately after the collision?

Accepted Answer

The conservation of momentum principle states that the total momentum before the collision is equal to the total momentum after the collision. Before the collision, the total momentum is $1200 \, \text{kg} \times 15.6 \, \text{m/s}$ (since the second car is stationary, it contributes no initial momentum). After the collision, the combined mass is $1200 \, \text{kg} + 1500 \, \text{kg} = 2700 \, \text{kg}$, and they move together with a velocity $v$. Setting the initial momentum equal to the final momentum, $1200 \times 15.6 = 2700 \times v$, solving for $v$ gives $v = 6.9 \, \text{m/s}$.

Question 17

A car heading north suddenly collides at an intersection with a truck of the same mass heading east.If they lock together and travel at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision,how fast was the car initially traveling?

Accepted Answer

The momentum of the system is conserved in both the x and y directions. In the x direction, the initial momentum is zero, and the final momentum is the mass of the two vehicles times their velocity in the x direction. In the y direction, the initial momentum is the mass of the car times its velocity in the y direction, and the final momentum is the mass of the two vehicles times their velocity in the y direction. Setting these two equations equal to each other and solving for the velocity of the car, we get:$$v_c = \frac{m_c v_{ce}}{m_c + m_t}$$where $m_c$ is the mass of the car, $v_{ce}$ is the velocity of the car in the x direction, $m_t$ is the mass of the truck, and $v_t$ is the velocity of the truck in the y direction.We are given that $m_c = m_t$, $v_{ce} = 28 \mathrm{~m/s}$, and $v_t = 15^\circ$. Substituting these values into the equation above, we get:$$v_c = \frac{(28 \mathrm{~m/s})\cos(15^\circ)}{2} = 14 \mathrm{~m/s}$$Therefore, the car was initially traveling at 14 m/s.

Question 18

A very small 51-g steel ball is released from rest and falls vertically onto a steel plate.The ball strikes the plate and is in contact with it for 0.50 ms.The ball rebounds elastically (at the same speed)and returns to its original height.The total time interval for a round trip is 3.00 s.What is the magnitude of the average force exerted on the ball by the plate during contact with the plate?

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Question 19

A block of mass m = 5.6 kg,moving on a frictionless surface with a velocity of $v _ { \mathrm { i } } = 6.5 \mathrm {~m} / \mathrm { s }$ to the right,collides with a block of mass M at rest,as shown in the figure.After the collision,the 5.6-kg block recoils with a velocity of $v _ { \mathrm { f } } = 0.70 \mathrm {~m} / \mathrm { s }$ to the left.If the blocks are in contact for 0.20 s,what is the magnitude of the average force exerted on the 5.6-kg block,while the two blocks are in contact?

Accepted Answer

First, we need to use conservation of momentum to find the velocity of the second block right after the collision:
\begin{align*}
m v_i &= m v_{1f} + M v_{2f} \
5.6 \cdot 6.5 &= 5.6 \cdot (-0.70) + Mv_{2f} \
Mv_{2f} &= 54.88 + 36.96 \
v_{2f} &= \frac{91.84}{M}
\end{align*}
Next, we can use the impulse-momentum theorem to find the average force:
\begin{align*}
F_{avg} &= \frac{\Delta p}{\Delta t} \
&= \frac{m\Delta v}{\Delta t} \
&= \frac{m(v_{1f} - v_i)}{\Delta t} \
&= \frac{5.6(-0.70 - 6.5)}{0.20} \
&= 202 	extrm{ N}
\end{align*}
Therefore, the answer is (A) 202 N.

Question 20

A block of mass m = 34 kg and speed V is behind a block of mass M = 81 kg and speed of $0.50 \mathrm {~m} / \mathrm { s }$ as shown in the figure.The surface is frictionless and the blocks collide and couple.After the collision,the blocks have a common speed of $0.90 \mathrm {~m} / \mathrm { s }$ What is the magnitude of the impulse on the 34-kg block due to the collision?

Accepted Answer

The answer of A block of mass m = 34...

Quiz 6: Impulse and Linear Momentum