Easy Impulse and Momentum Problems for 9th Graders

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In a billiards game, one player hits the cue ball towards another ball. The cue ball has a mass of 0.1kg and hits the other ball with a velocity of $2\frac{m}{s}$ . If the collision is completely elastic and the cue ball travels with a velocity of $0.8\frac{m}{s}$ after the collision, what is the mass and velocity of the other ball?

Correct answer:

$2.8\frac{m}{s},\ 0.043kg$

Explanation:

Since the collision is completely elastic, we know that both momentum and kinetic energy are conserved. We can write the following equations (initial momentum and energy of the second ball are neglected since it is not moving:

m_1v_i=m_1v_f +m_2v_2

$\frac{1}{2}m_1v_i^2= \frac{1}{2}m_1v_f^2+\frac{1}{2}m_2v_2^2$

Rearrange the first equation for m_2 and the second for v_2 .

$m_2=\frac{m_1(v_i-v_f)}{v_2}$

$v_2^2=\frac{m_1(v_i^2-v_f^2)}{m_2}$

We can rewrite the second equation as:

$v_2^2=\frac{m_1(v_i+v_f)(v_i-v_f)}{m_2}$

Substitute our equation for m_2 into the second equation:

$v_2^2=\frac{m_1(v_i-v_f)(v_i+v_f)}{\frac{m_1(v_i-v_f)}{v_2}}$

Rearranging, we get:

$v_2^2=\frac{m_1(v_i-v_f)(v_i+v_f)}{m_1(v_i-v_f)v_2^{-1}}=\frac{(v_i+v_f)}{v_2^{-1}}$

v_2^2=v_2(v_i+v_f)

v_2 = (v_i +v_f)

Plug in our values for the initial and final velocities:

$v_2=(2\frac{m}{s}+0.8\frac{m}{s})$

$v_2=2.8\frac{m}{s}$

To solve for mass, we'll use our earlier expression for m_2 :

$m_2=\frac{m_1(v_i-v_f)}{v_2}$

$m_2= \frac{(0.1kg)(2\frac{m}{s}-0.8\frac{m}{s})}{2.8\frac{m}{s}}= 0.043kg$

Two sumo wrestlers are in a match. At the start of the match, they both lunge at each other. They hit and miraculously come to a stand still. One wrestler was 200kg and traveling at a velocity of $2.3\frac{m}{s}$ at the instance of collision. If the other wrestler was traveling at $2.9\frac{m}{s}$ , what is his mass?

Correct answer:

159 kg

Explanation:

It does not matter whether the collision is elastic or inelastic (although it would be best to assume that it's inelastic). Momentum is conserved in either type of collision, and is the only value needed for our calculation. Since they come to a standstill, their momentums at the moment of collision are equal and opposite:

m_1v_1=m_2v_2

Rearrange to solve for m_1 :

$m_1=\frac{m_2v_2}{v_1}$

Plug in the given values from the question and solve:

$m_1=\frac{(200kg)(2.3\frac{m}{s})}{2.9\frac{m}{s}}$

m_1= 159 kg

One car with a mass of 400kg is traveling east at $20\frac{m}{s}$ and collides with a car of mass 800kg traveling west at $15\frac{m}{s}$ . Assuming the collision is completely inellastic, what is velocity of the first car after the collision?

Correct answer:

$3.33\frac{m}{s}\ \text{west}$

Explanation:

Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:

p_i = p_f

There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:

m_1v_1 + m_2v_2 = mv

Rearranging for final velocity, we get:

$v = \frac{m_1v_1 + m_2v_2}{m}$

At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:

$v = \frac{(400)(-20)+ (800)(15)}{400 + 800}$

$v = \frac{-800 + 12000}{1200} = \frac{4000}{1200} = 3.33\frac{m}{s}$

Since this value is positive, the final answer is $3.33\frac{m}{s}$ West.

A marble of mass 0.1kg falls off a bed with a height of 1.2 m . What is the impulse on the marble as it hits the ground?

$g = 10\frac{m}{s^2}$

Correct answer:

$0.49\ N\cdot s$

Explanation:

Impulse can be written as either of two popular expressions:

$I = F\Delta t = m\Delta v$

From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:

E=U_i+K_i = U_f + K_f

Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:

$mgh_i = \frac{1}{2}mv_f^2$

Canceling out mass and rearranging for final velocity, we get:

$v_f = \sqrt{2gh_i}$

We know these variables, allowing us to solve for the velocity:

$v_f = \sqrt{2(10\frac{m}{s^2})(1.2m)} = 4.9\frac{m}{s}$

Plugging this value into the expression for impulse, we get:

$I = (0.1kg)(4.9\frac{m}{s})= 0.49 N\cdot s$

Consider the following system:

Slope_2

If the block has a mass of 20kg , the angle measures $50^{\circ}$ , and there is no friction between the block and slope, what is the momentum of the block after it has traveled a horizontal distance of 10m ?

Correct answer:

$309\frac{kg\cdot m}{s}$

Explanation:

To calculate the momentum of the block, we first need to know the velocity of the block. This can be found using the equation for the conservation of momentum:

E = U_i+K_i = U_f +K_f

If we assume that the final height is zero, we can eliminate initial kinetic energy and final potential energy, getting:

U_i = K_f

Substituting expressions for each term, we get:

$mgh_i = \frac{1}{2}mv_f^2$

Cancel out mass and rearrange to solve for velocity:

$v_f = \sqrt{2gh_i} = \sqrt{2(10\frac{m}{s^2})(h_i)}$

We can use the horizontal distance traveled and the angle of the slope to determine the initial height:

$tan(A) = \frac{h_i}{horizontal}$

$h_i = tan(A)\cdot horizontal$

$h_i = tan(50^{\circ})*10m = 11.9m$

Now that we have the initial height, we can solve for final velocity:

$v_f = \sqrt{2(10\frac{m}{s^2})(11.9m)} = 15.4\frac{m}{s}$

Finally, we can now use the equation for momentum to solve the problem:

$p = m\Delta v$

$p = (20kg)(15.4\frac{m}{s})$

$p = 309 \frac{kg\cdot m}{s}$

Two astronauts in space are traveling directly towards each other. Astronaut A has a mass of 70kg and a velocity of $15\frac{m}{s}$ and Astronaut B has a mass of 60kg and a velocity of $5\frac{m}{s}$ . When the astronauts collide, they grab onto each other. What is the velocity of the two astronauts after the collision as they continue to grab onto each other?

Correct answer:

$5.77\frac{m}{s}$

Explanation:

Momentum is always conserved. Equation for conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{total}$

There is only one velocity on the right since the two astronauts grab onto each other, thus they move together at the same velocity. Solve.

$(70kg\cdot15\frac{m}{s})+(60kg(-5\frac{m}{s}))=(70kg+60kg)v_{total}$

$v_{total}=5.77\frac{m}{s}$

A 5 kg rock flying through the air is traveling at a velocity of $12\frac{m}{s}$ when it collides into and sticks to a stationary bean bag, weighing 20kg . What is the velocity of the two objects?

Correct answer:

$2.4\frac{m}{s}$

Explanation:

The equation for momentum is:

p=mv

To maintain conservation of momentum, a new state must have the same momentum as a previous state:

$m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1f}v_{1f}+m_{2f}v_{2f}$

Since the rock and the bean bag move together after the collision, $v_{1f}=v_{2f}$ . And since the bean bag is initially stationary, $v_{2i}=0$

Plug in known values and solve.

$5kg(12\frac{m}{s})+20kg(0)=(5kg+20kg)(v_f)$

$60\frac{kgm}{s}=25kg(v_f)$

$v_f=2.4\frac{m}{s}$

A 150g baseball is thrown with a speed of $40\: \frac{m}{s}$ . If it takes 0.7s for the baseball to come to rest in the catcher's glove, what is the average force the catcher experiences due to the ball?

Correct answer:

8.57N

Explanation:

To solve this problem, we need to consider the change in the ball's momentum. To do so, we'll use the following equation.

$J=m\Delta v=F_{avg}t$

Rearrange the above equation to solve for the average force.

$F_{avg}=\frac{m\Delta v}{t}=\frac{(0.15 kg)(40 \frac{m}{s})}{0.7 s}$

$F_{avg}=8.57N$

Joe, of mass 90kg, jumps straight up. To do so, he bends his knees and produces an upwards force that results in a constant upward net force of 100N. If Joe experiences this force for 0.9s before leaving the ground, what is Joe's velocity immediately after he leaves the ground?

Correct answer:

$1\frac{m}{s}$

Explanation:

To solve this problem we need to use the relationship between force and impulse, which is given by the following equation:

$F=\frac{\Delta P}{t}$

This equation represents that the rate of change of momentum with respect to time is equal to the net force that causes said change in momentum. Thus:

$F=\frac{P_f-P_i}{t}=\frac{mV_f-mV_i}{t}$

Note that Joe must have an initial velocity of $0\frac{m}{s}$ before he begins to apply the upwards Force that accelerates him upwards, therefore our equation simplifies to:

$F=\frac{mV_f}{t}$

Solve for V_f :

$V_f=\frac{Ft}{m}=\frac{(100N)(0.9s)}{90kg}=\frac{90kg\cdot \frac{m}{s}}{90kg}=1\frac{m}{s}$

Which of the following explains why when we land on our feet, we instinctively bend our knees? Hint: think about the relationship between force, impulse, and time.

Possible Answers:

By bending our knees we extend the time it takes us to stop, which increases the impact force

When we bend our knees we extend the time in which we apply the force that stops us, so our impulse is greater

By bending our knees we extend the time it takes us to stop, which diminishes the impact force

When we bend our knees we extend the time in which we apply the force that stops us, so our impulse is smaller

By bending our knees we use a greater force to stop, which makes the impulse smaller

Correct answer:

By bending our knees we extend the time it takes us to stop, which diminishes the impact force

Explanation:

Say that, when we hit the ground, we have a velocity, which is predetermined by whatever happens before the impact. When we hit the ground you will experience a force for some time. This force will cause the acceleration that reduces our velocity to zero and gets us to stop. Note that, regardless of how much time it takes us to stop, the change in momentum (impulse) is fixed, since it directly depends on how much our velocity changes:

P_i=mv

$P_f=\frac{0kg\cdot m}{s}$ (since we come to a stop)

Note that the initial momentum does not depend on the impact force nor on how much time it takes to stop. The initial momentum depends on the velocity we have when we first hit the ground. This velocity is given by whatever happened before we hit the ground, which no longer concerns us since we only care about what happens from the moment we first hit the ground till the moment we stop. Yes, the time that passes for you to stop is very small, but it is impossible for it to be zero. So we have that the change in momentum (impulse) is a constant:

$\Delta P=-mv= constant$ , since is predetermined.

Remember that any change in momentum for a given mass occurs because its velocity changes. The velocity of the mass changes due to an acceleration and an acceleration is caused by a force. This gives us a relationship between force and impulse:

$F=\Delt\frac{\Delta P}{t}$

In our scenario, would be the impact force that stops us and the time it takes us to stop. From the equation above, it is easy to see that, since $\Delta P$ is fixed, when gets larger gets smaller, and the other way around. Therefore, we bend our knees to effectively increase the time it takes us to stop. Thus, diminishing the impact force as to avoid hurting ourselves.

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Easy Impulse and Momentum Problems for 9th Graders

All AP Physics 1 Resources

All AP Physics 1 Resources

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