A 1.7 kg ball strikes a wall with a velocity of 7.7 m/s to the left. The ball bounces off with a velocity of 6.5 m/s to the right. If the ball is in contact with the wall for 0.17 s, what is the constant force exerted on the ball by the wall?
force*time=changeinmomentum=1.7kg*(6.5-(-7.7)) ?
solve for force.
To solve this problem, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
First, let's calculate the initial momentum of the ball before the collision. The formula for momentum is given by:
momentum = mass * velocity
The mass of the ball is 1.7 kg, and its initial velocity is 7.7 m/s to the left. Therefore, the initial momentum is:
momentum_before = mass * velocity_before = 1.7 kg * (-7.7 m/s) = -13.09 kg * m/s
The negative sign indicates the direction of the velocity.
Next, let's calculate the final momentum of the ball after the collision. The mass of the ball remains the same at 1.7 kg, but its velocity is now 6.5 m/s to the right. Therefore, the final momentum is:
momentum_after = mass * velocity_after = 1.7 kg * 6.5 m/s = 11.05 kg * m/s
The difference between the initial and final momenta represents the change in momentum of the ball. According to Newton's second law of motion, this change in momentum is equal to the force applied on the ball multiplied by the contact time.
change_in_momentum = force * contact_time
We are given the contact time as 0.17 s. Therefore, we can rearrange the equation to solve for the constant force exerted on the ball:
force = change_in_momentum / contact_time
Substituting the values we calculated:
force = (momentum_after - momentum_before) / contact_time
= (11.05 kg * m/s - (-13.09 kg * m/s)) / 0.17 s
= (11.05 kg * m/s + 13.09 kg * m/s) / 0.17 s
= 24.14 kg * m/s / 0.17 s
≈ 142.00 N
Therefore, the constant force exerted on the ball by the wall is approximately 142.00 N.