Post

AP Physics 1 Exam: Momentum and Impulse Problem

Created by @nathanedwards
 at November 3rd 2023, 9:39:48 am.
AP_Physics_1-Questions
#momentum
#collisions
#ap-physics-1

AP Physics 1 Exam Question

A 0.2 kg baseball traveling horizontally with a velocity of 15 m/s collides with a stationary 0.5 kg baseball bat. The collision is completely elastic, and after the collision, the baseball bounces back with a velocity of 10 m/s in the opposite direction.

a) Calculate the momentum of the baseball before the collision. b) Calculate the momentum of the baseball bat after the collision. c) Determine the impulse experienced by the baseball bat during the collision. d) Find the average force exerted on the baseball bat during the collision.

Answer:

a) To calculate the momentum of the baseball before the collision, we use the formula:

p=mv p = mv

where p p is the momentum, m m is the mass of the baseball, and v v is the velocity of the baseball.

Given: Mass of the baseball, m1=0.2 m_1 = 0.2 kg Velocity of the baseball, v1=15 v_1 = 15 m/s

Substituting the values into the formula, we get:

p=(0.2 kg)(15 m/s)=3 kg\cdotpm/s p = (0.2 \, \text{kg})(15 \, \text{m/s}) = 3 \, \text{kg·m/s}

Therefore, the momentum of the baseball before the collision is 3 kg·m/s.

b) Since the collision is elastic, the momentum of the baseball bat after the collision can be expressed as:

p=−mv p = -mv

where m m is the mass of the baseball bat (0.5 kg), and v v is the velocity of the baseball (10 m/s).

Substituting the values, we get:

p=−(0.5 kg)(10 m/s)=−5 kg\cdotpm/s p = -(0.5 \, \text{kg})(10 \, \text{m/s}) = -5 \, \text{kg·m/s}

Therefore, the momentum of the baseball bat after the collision is -5 kg·m/s.

c) The impulse experienced by the baseball bat during the collision can be calculated using the formula:

Impulse=Δp \text{Impulse} = \Delta p

where Δp \Delta p is the change in momentum.

Given: Initial momentum of the baseball bat, pinitial=0 p_{\text{initial}} = 0 kg·m/s Final momentum of the baseball bat, pfinal=−5 p_{\text{final}} = -5 kg·m/s

Substituting the values into the formula, we get:

Impulse=pfinal−pinitial=(−5 kg\cdotpm/s)−(0 kg\cdotpm/s)=−5 kg\cdotpm/s \text{Impulse} = p_{\text{final}} - p_{\text{initial}} = (-5 \, \text{kg·m/s}) - (0 \, \text{kg·m/s}) = -5 \, \text{kg·m/s}

Therefore, the impulse experienced by the baseball bat during the collision is -5 kg·m/s.

d) The average force exerted on the baseball bat during the collision can be calculated using the formula:

Impulse=F⋅Δt \text{Impulse} = F \cdot \Delta t

where Impulse \text{Impulse} is the impulse calculated in part (c), F F is the average force, and Δt \Delta t is the duration of the collision.

We need to rearrange the formula to solve for F F :

F=ImpulseΔt F = \frac{\text{Impulse}}{\Delta t}

Given: Impulse, Impulse=−5 \text{Impulse} = -5 kg·m/s Duration of the collision, Δt \Delta t (not provided)

Since the duration of the collision is not provided, we cannot calculate the exact average force without this information.

End of answer.

Chat

Save To Bookmark