Physics, asked by dhruvjainip94ppw, 1 year ago

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

Answered by YASHOFFICIAL24082004
14
F = dP/dt = m*dv/dt = .1*(5-4)/.15 = .666 N
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
Answered by soniatiwari214
0

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.

#SPJ2

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