rce versus time during the collision of a 44- tennis ball with a wall. The initial velocity of the ball is 36 perpendicular to t
he wall; it rebounds with the same speed, also perpendicular to the wall. What is the maximum value of the contact force during the collision
1 answer:
Answer:
Maximum contact force = 487.4 N
Explanation:
The graph showing force vs time is attached to the solution of this question.
According to the an interpretation of Newton's second law of motion, the magmitude of the change in momentum is equal to the magnitude of the impulse experienced duting this impact.
F.t = Δ(momemtum)
Δ(momemtum) = 0.044 [36 - (-36)] = 0.044 × 72 = 3.168 kgm/s
Meaning, Impulse = 3.168 kgm/s too.
The area under the curve above is the impulse we just obtained.
If the maximum contact force is called F
The impulse = Area under this graph
= (1/2)(F)(4-0) + (F)(7-4) + (1/2)(F)(10-7)
= 2F + 3F + 1.5F = (6.5F) mN.s = (0.0065F) N.s
0.0065F = 3.168
F = (3.168/0.0065) = 487.4 N
Hope this Helps!!!
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Answer:
The total momentum is zero.
Explanation:
This problem can be solved by applying the momentum conservation theorem and the amount of motion. This theorem tells us that the amount of motion is conserved before and after a collision.
In the next equation, we will write to the left of the equal sign the amount of motion before the collision and to the right the amount of motion after the collision.
where:
P₁ = momentum of the ball moving to the right, before the collision = 85 [kg*m/s]
P₂ = momentum of the ball moving to the left, before the collision = - 85 [kg*m/s]
P₃ = Final momentum after the collision [kg*m/s]
There is no movement of any of the balls, they remain at rest after the impact.