A 0.10 kg ball is dropped onto a table top. The speeds of the ball right before and right after hitting the table top are 5.0 m/s and 4.0 m/s, respectively. If the collision between the ball and the table top last 0.15 s, what is the magnitude of the average force exerted on the ball by the table top
Answers
If you meant that the ball bounces upward @ 4 m/s, then V2 = -4 m/s and
F = .1*(5+4)/.15 = 6.0 N
Concept:
The relationship between mass and velocity is known as momentum. It is expressed as P = mv
Given:
Mass of ball = 0.10 kg
The speeds of the ball right before and after are 5.0 m/s and 4.0 m/s, respectively.
Collision time = 0.15seconds
Find:
We need to determine the magnitude of the tabletop's average force exerted on the ball.
Solution:
The relationship between mass and velocity is known as momentum. It is expressed as P = mv where P is the momentum, m is the mass and v is the velocity
Change in the ball's momentum equals the difference between the final and initial momentum.
Thus, Change in the momentum of the ball = final momentum, P2 - Initial momentum, P1
Final momentum, P2 = mass × Final velocity
P2 = 0.1 × 4 = 0.4 kg.m/s
Initial momentum, P1 = mass × Initial velocity
P1 = 0.1 × (-5.0) = -0.5 kg.m/s
(taking downward velocity as negative, -ve).
We know, Force = dP/dt
Force = (P2 - P1)/dt
Force = [0.4 - (-0.5)]/0.15
Force = 0.90/0.15
F = 6.0 N
Thus, the magnitude of the average force exerted on the ball by the tabletop is 6.0 N.
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