A tennis ball of mass m = 0.060 kg is traveling with a speed of 28 m/s and θ = 35 degrees. It strikes a wall and bounces off wit
h a speed of 28 m/s. What is the magnitude of the force, in Newtons, exerted on the ball if the collision time is 0.1 s? what direction was the force exerted by the wall?
1 answer:
Answer:
19.15 Newton
Explanation:
Force on a body is rate of change of momentum.
Thus force on wall is =
ΔP = P(final) - P(initial) [in vector]
Note : take component of velocity normal to wall (assuming wall smooth)
P(final) = m*v*sin(Ф) [in negative x direction] = -0.06*28*sin(35)
P(initial) = m*v*sin(Ф) [in positive x direction] = 0.06*28*sin(35)
Thus ΔP=2(0.06*28*sin(35))
Now ΔT is given =0.1 s
Force on wall = =
=19.15 Newton
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Answer:
a) 45571 N
b) 22786 N
c) 4557 N
Explanation:
- Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
- In other words, the change in momentum, must be equal to the initial one, but with opposite sign.
- Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:
- So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows: