Question

A ball 10 kg hits a hard surface at 60°to normal speed 5m/s and rebounds with 3m/s at same angle if bal remains in contact for 0.1 seconds ,what force does it
exert

Answer 1

Given,

Mass of the ball = 10 kg

Speed of ball = 5 m/s

Speed of the ball after rebounding = 3 m/s

Time taken = 0.1 sec

To Find,

Force exerted by the ball.

Solution,

The formula for calculating the force when momentum is involved is Force= ΔP/Δt, where P is momentum

         = m(v2-v1)/0.0 , where mis thee  mass of the ball of in kg

         = 0.1/0.01

         = 10 N

Hence, the force exerted by the ball is 10 N.

Answer 2

Answer:

The force exerted by the ball is 400N.

Explanation:

From question we have,

m=10kg

θ=60°

v₁=5m/s

v₂=3m/s

t=0.1seconds

During the rebound, the horizontal component ( parallel to the surface ) of velocity does not change but the vertical component(perpendicular to the surface) of velocity changes.

$\Delta P_x=0\\\Delta P_Y\neq0$                (1)

ΔPx=change in momentum in the horizontal direction

ΔPy=change in momentum in the vertical direction

So,

$\Delta P_y=mv_2cos 60\textdegree-(-mv_1cos 60\textdegree)$         (2)

$\Delta P_y=10\times 3\times \frac{1}{2}+10\times 5\times \frac{1}{2}=40$           (3)

Now the force is given as,

$F=\frac{\Delta P}{t}$        (4)

ΔP=change in momentum

t=time in which the change has occurred

F=force

By substituting the value of ΔP and t in equation (4);

$F=\frac{40}{0.1}=400N$

Hence, the force exerted by the ball is 400N.