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AP Physics 1 : Impulse and Momentum

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Example Question #461 : Newtonian Mechanics

Deep in space Object has mass 350kg and is initially traveling with velocity $<30,50>\frac{m}{s}$ . At t=0 , it collides with Object , which has mass 500kg and is initially motionless. The two objects stick together.

Determine the initial momentum of the system.

Possible Answers:

$<10500,17500>\frac{kg*m}{s}$

Impossible to determine

$<17500,17500>\frac{kg*m}{s}$

None of these

$<17500,10500>\frac{kg*m}{s}$

Correct answer:

$<10500,17500>\frac{kg*m}{s}$

Explanation:

Using

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

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

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Example Question #462 : Newtonian Mechanics

Determine the impulse experienced by Object

Possible Answers:

None of these

$<10300,6200>\frac{kg*m}{s}$

$<4340,7210>\frac{kg*m}{s}$

$<4449,39080>\frac{kg*m}{s}$

$<6200,10300>\frac{kg*m}{s}$

Correct answer:

$<6200,10300>\frac{kg*m}{s}$

Explanation:

Using

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

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

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

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v_f}=<12.4,20.6>\frac{m}{s}$

Definition of impulse:

$J=\Delta \overrightarrow{p}$

$J=m\Delta\overrightarrow{v}$

J=500(<12.4,20.6>-<0,0>)

$J=<6200,10300>\frac{kg*m}{s}$

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Example Question #41 : Impulse And Momentum

A ball of mass 30g is thrown at a target. The ball strikes with a velocity of $20\frac{m}{s}$ and bounces back with equal magnitude. Determine the magnitude of impulse experienced by the ball.

Possible Answers:

$5.5\frac{kg\cdot m}{s}$

$6.8\frac{kg\cdot m}{s}$

$7.7\frac{kg\cdot m}{s}$

$1.2\frac{kg\cdot m}{s}$

$3.6\frac{kg\cdot m}{s}$

Correct answer:

$1.2\frac{kg\cdot m}{s}$

Explanation:

Impulse is defined as change in momentum:

$\Delta P= P_f-P_i$

Using

P=mv

Combining equations:

$\Delta P= mv_f-mv_i$

Plugging in values:

$\Delta P= .030*-20-.030*20$

$|\Delta P| =1.2\frac{kg*m}{s}$

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Example Question #41 : Impulse And Momentum

Frictionless cart is traveling at $<10,15>\frac{m}{s}$ when it hits identical frictionless cart which was previously motionless. After the collision, cart is traveling at $<1,1.5>\frac{m}{s}$ . Determine the final velocity of cart .

Possible Answers:

$<18,27>\frac{m}{s}$

$<9,13.5>\frac{m}{s}$

$<0,0>\frac{m}{s^2}$

None of these

$<19,3.5>\frac{m}{s}$

Correct answer:

$<9,13.5>\frac{m}{s}$

Explanation:

Using conservation of momentum:

Pi=P_f

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Solving for $v_{2f}$

$\frac{m_1v_{1i}+m_2v_{2i}-m_1v_{1f}}{m_1}=m_2v_{2f}$

Using m_1=m_2 (since the carts are identical):

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

Plugging in values:

<10,15>+<0,0>-<1,1.5>=<9,13.5>

$v_{2f}=<9,13.5>\frac{m}{s}$

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Example Question #41 : Impulse And Momentum

During time period , a rocket ship deep in space of mass 2.55*10^5kg travels from <0,10>km to <5,5>km . During time period , the rocket fires. During time period , the rocket travels from <-15,0>km to <-30,-5>km .

Time periods , , and all took 10seconds.

Determine the impulse during time period .

Possible Answers:

$\Delta\overrightarrow{P}=<0,5.1*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<5.1*10^8,5.1*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<0,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

None of these

$<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

$<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Explanation:

Finding initial momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P_i}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Converting to and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}_f=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

Using

$Impulse=\Delta\overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}$ Plugging in values:

$\overrightarrow{\Delta P} =<-3.825*10^8,-1.275*10^8>-<1.275*10^8,-1.275*10^8>$

$\Delta\overrightarrow{P}=<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

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Example Question #42 : Impulse And Momentum

Ty throws a 0.10kg snowball directly at his brother who also throws a snowball directly at Ty with a mass of 0.20kg , because he dipped it in a bucket of water. Ty throws his snowball at $20.0\frac{m}{s}$ and his brother throws his at $-15\frac{m}{s}$ and the two snowballs make perfect contact in mid flight, causing a perfectly inelastic collision. Neglecting air resistance, what is the combined final speed of the snowballs and their direction following the collision?

Possible Answers:

$6.6\frac{m}{s}$ ; positive x-direction

$3.3\frac{m}{s}$ ; positive x-direction

$-3.3\frac{m}{s}$ ; negative x-direction

$-6.6\frac{m}{s}$ ; negative x-direction

Correct answer:

$-3.3\frac{m}{s}$ ; negative x-direction

Explanation:

Use the conservation of momentum for inelastic collisions (objects that stick together).

$m_{Ty}(v_{Ty})+m_{brother}(v_{brother})=(m_{Ty}+m_{brother})v_{final}$

Plug in and solve for the final velocity.

$0.1Kg(20\frac{m}{s})+0.2Kg(-15\frac{m}{s})=(0.1Kg+0.2Kg)v_{final}$

$2j+(-3j)=(0.3Kg)v_{final}$

$v_{final}=-3.33\frac{m}{s}$

The final velocity of the combined snowballs is $-3.3\frac{m}{s}$ in the negative x-direction.

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Example Question #467 : Newtonian Mechanics

On a toy car set, two children apply a piece of double sided tape to the front of two toy cars so that when they collide they will stick together causing a perfectly inelastic collision. The two identical cars have a combined mass of 0.075kg and their final combined speed was $2.60\frac{m}{s}$ in the rightward direction. If car "A" was shot to the right with a speed measured at $17.0\frac{m}{s}$ right before the collision . At what speed was the other car traveling right before the collision?

Possible Answers:

$-24.5\frac{m}{s}$

$-18.4\frac{m}{s}$

$-11.8\frac{m}{s}$

$-6.1\frac{m}{s}$

Correct answer:

$-11.8\frac{m}{s}$

Explanation:

Use the conservation of momentum, perfectly inelastic collision.

$m_{A}(v_{A})+m_{B}(v_{B})=(m_{A}+m_{B})v_{final}$

Plug in and solve for the speed of car B.

$0.0375Kg(17\frac{m}{s})+0.0375Kg(v_{B})=(0.075Kg)2.6\frac{m}{s}$

$0.6375j+0.0375Kg(v_{B})=0.19j$

$v_{B}=-11.8\frac{m}{s}$

The velocity of car B right before the collision was $-11.8\frac{m}{s}$ .

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Example Question #43 : Impulse And Momentum

A spaceship is stationary deep in space with it’s rocket broken. The crew decides to propel it to earth by throwing tennis balls out the back window. If they throw one tennis ball every second at $30 \frac{m}{s}$ , how many balls will it take the ship to reach $1.1*10^4 \frac{m}{s}$ ? Each tennis ball has a mass of 50 g and the ship has a mass of 7.0*10^6kg .

Possible Answers:

$6.26*10^{14}balls$

$5.13*10^{10}balls$

$8.44*10^{7}balls$

$3.95*10^{18}balls$

None of these

Correct answer:

$5.13*10^{10}balls$

Explanation:

The net momentum of the system needs to stay constant, that is, any momentum given to the tennis balls, the space ship must gain in the opposite direction.

m_sv_s+n_bm_bv_b=0

Where

m_s is the mass of the ship.

v_s is the velocity of the ship

n_b is the number of balls thrown

m_b is the mass of a ball

v_b is the velocity of a thrown ball

Solving for

$n_b=\frac{m_sv_s}{m_b(-v_b)}$

Since the velocities are given as magnitudes, the negative sign can be ignored

$n_b=\frac{m_sv_s}{m_b(v_b)}$

Plugging in values

$n_b=\frac{7*10^6*1.1*10^4}{.050(30)}$

$n_b=5.13*10^{10}$

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Example Question #501 : Ap Physics 1

A bullet with mass is shot with an initial kinetic energy of 2.5kJ . What was the impulse on the bullet as it was fired?

Possible Answers:

$5N\cdot s$

$25N\cdot s$

$10N\cdot s$

$20N\cdot s$

$15N\cdot s$

Correct answer:

$5N\cdot s$

Explanation:

The expression for impulse is the following:

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

We are given mass and initial kinetic energy, so we can also calculate the velocity of the bullet. Therefore, we will go with the second for of the expression:

$I = m\Delta v$

Finding an expression for velocity, we get:

$K= \frac{1}{2}mv^2$

Rearranging for velocity, we get:

$v = \sqrt{\frac{2K}{m}}$

Plugging this in, we get:
$I = m\sqrt{\frac{2K}{m}}=\sqrt{2Km}$

Plugging in our values, we get:

$I = 5N\cdot s$

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Example Question #41 : Impulse And Momentum

A bullet of mass is shot at a velocity of $950\frac{m}{s}$ and an angle of $35^{\circ}$ above the horizontal. What is the momentum of the bullet at its maximum height? Neglect air resistance.

Possible Answers:

$0.4\frac{kg\cdot m}{s}$

$5.0\frac{kg\cdot m}{s}$

$3.2\frac{kg\cdot m}{s}$

$1.9\frac{kg\cdot m}{s}$

$4.1\frac{kg\cdot m}{s}$

Correct answer:

$4.1\frac{kg\cdot m}{s}$

Explanation:

When the bullet is at its maximum height, it has no vertical velocity component. Also, since we are neglecting air resistance, the horizontal velocity remains constant. Therefore, the velocity of the bullet at its maximum height is simply the initial horizontal velocity:

$v_x = v_icos(\theta)$

Plugging in values, we get:

$v_x = \left ( 950\frac{m}{s}\right )cos(30^{\circ})$

$v_x = 822.7\frac{m}{s}$

Then using the expression for momentum:

p=mv

$p = (0.005kg)\left ( 822.7\frac{m}{s}\right )$

$p = 4.1\frac{kg\cdot m}{s}$

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AP Physics 1 : Impulse and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #461 : Newtonian Mechanics

Example Question #462 : Newtonian Mechanics

Example Question #41 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

Example Question #42 : Impulse And Momentum

Example Question #467 : Newtonian Mechanics

Example Question #43 : Impulse And Momentum

Example Question #501 : Ap Physics 1

Example Question #41 : Impulse And Momentum

All AP Physics 1 Resources

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