Physics, asked by fnaticabhinavtx, 11 months ago

A foot ball of mass 0.5 kg moving with a velocity
of 10 m s-1 hits a pole and, the ball turns back and
moves with a velocity of 20 m s-1. Find the impulse
and the force exerted on the ball if the force acts for
0.02 s.​

Answers

Answered by shadowsabers03
12

Given,

m = 0.5 kg

t = 0.02 s

u = 10 m s^(-1)

After collision with the pole, the ball turns back so the velocity will be in opposite direction. Hence,

v = - 20 m s^(-1)

We know that impulse of a body is the change in momentum of that body. So,

Impulse = Final momentum - Initial momentum, i.e.,

I = m (v - u)

I = 0.5 (- 20 - 10)

I = 0.5 × (- 30)

I = - 15 kg m s^(-1)

And the force exerted on the ball is the rate of change of momentum, or the impulse imparted per unit time, i.e.,

F = I / t

F = - 15 / 0.02

F = - 750 N

Answered by Ranveerx107
40

Answer

Momentum of ball is changed by 7.5 kg m/s.

A force of 375 N is exerted on the ball.

Explanation

Given:-

Mass of football, m = 0.5 kg

Initial velocity of football, u = 10 m/s

Ball hits a pole and bounces back with certain velocity

Final velocity of football, v = -5 m/s   [Negative because ball bounces back]

time for which force acts, t = 0.02 s

To find:-

Change in momentum of the ball, Δp =?

Force exerted on the ball, F =?

Formulae required:-

Formula for Change in momentum

        Δp = (Final p) - (Initial p) = m v - m u

Formula for force

        F = Δp / t

Where, p is momentum, m is mass, u is initial velocity, v is final velocity, t is time.

Solution:-

Calculating the change in momentum of ball

→ Δp = (Final p) - (Initial p) = m v - m u

→ Δp = m v - m u

→ Δp = m ( v - u )

→ Δ p = 0.5 ( -5 -10 ) = -7.5 kg m/s

Therefore,

The magnitude of change in momentum of the ball is 7.5 kg m/s.

Now, Calculating force exerted on the ball

→ F = Δp / t

→ F = (-7.5) / 0.02

→ F = -375 N

Therefore,

A force of magnitude 375 N is exerted on the ball by the wall.

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