# A 2.5 kg ball strikes a wall with a velocity of 8.2 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 0.22 s, what is the...

A 2.5 kg ball strikes a wall with a velocity of 8.2 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 0.22 s, what is the constant force exerted on the ball by the wall? Answer in units of N.

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### 3 Answers

A ball and a lump of plasticine of equal mass are thrown at a wall. The ball is in contact with the wall for 0.05 s it then bounces back. The plasticine hits the wall and doesn't bounce back. Explain why the ball exerts a larger force on the wall then the plasticine.

Initial velocity of ball = u = 8.2 m/s

Finial velocity of ball = v = -6.5 m/s

Time take for the ball to change velocity = time the ball is in contact with the wall = t = 0.22 s

therefore acceleration of ball while being bounced off the wall

= a = (v-u)/t = (-6.6-8.2)/0.22 = m?s^2

And force = f = mass of ball*a = 2.5*[(-14.8)/0.22] = 168.1818 N (approximately)

Answer: constant force exerted by the wall is 168.1818 N.

The force is defined as the rate of change in momentum of the body.

The momentum of the ball is the product of its mass and velocity and it is a vector.

So the momentum of the ball at the time it strike the wall = 2.5 kg*8.2m/s = 20.5kg.m/s left.

The monteum of the ball when it is reflected by the wall = 2.5kg*6.5m/s right =16.25 kg.m/s right=-16.25 kg.m/s considering the left direction.

Therefore, the change in momentum =20.5-(-16.25) =36.75 kg*m/s

Threfore the rate of change of momentum ={36.75kg*m/s}/Time of contact={36.75kg*m/s}/0.22s =167.0455 N is the force exerted by the wall.