Question

hockey ball at rest is it by hockey stick such that the force acts on the ball for 0.08 seconds if the ball is of mass hundred grams and covers a distance of 80 m in 1.6 seconds find the magnitude of the force

Answer 1

The ball covers a distance of 80 m in 1.6 seconds.

Velocity of ball after it was hit, v = distance/ time = 80/1.6 = 50 m/s

Velocity Before it was hit, u = 0

Mass of ball, m = 100g = 0.1 kg

Change in momentum of ball due to hitting, ∆P = m(v-u) = 0.1×(50-0) = 5 Ns

Time for which the force acted, t = 0.08s

We know that ∆P = Ft

5 = F × 0.08

F = 5 ÷ 0.08

F = 62.5 N

The force is 62.5 N

Answer 2

Answer:p =62.5 N

Explanation:

Distance covered by the ball when force is acting on it = 80m

So, Uniform velocity of the ball = d/t

=80m/1.6s = 50 ms-

So,final velocity of the ball after the force stops acting on it = 50 ms-

Initial velocity of the ball =0

Time for which the force acts = 0.08s

Applying, v=u+at

50m- = 0+ a × 0.08s

Or, a=( 50/0.08) ms-^2

So, force acting on the ball.....

= m× a =(100/1000)kg × 625 ms-^2

=62.5N