Question

A ball of mass 50kg is moving horizontal with a speed of 20 m/s . a man hits the ball and ball travels back in opposite direction with the same speed . if ball is contact with the bat for 0.1 s, the force exerted by bat on ball is

Answer 1

2×10000 N
bcoz change in momentum is 50 ×2 ×20 kgms^-1
(as in opposite direction)
and force is change in momentum over time.
i.e. 2000÷0.1=20000N

Answer 2

Answer:

$\large \tt \ \: {force = \frac{change \: in \: momentum}{time} }$

$\large \tt \ \: {force = \frac{m \: delta \: v }{time} }$

$\implies \: \large\tt \ \: { \frac{m |v - u| }{t} }$

$\implies \: \frac{50 |20 - ( - 20)| }{0.1}\\ \\ \implies \: \frac{ 50 |20 + 20| }{0.1} \\ \\ \implies \: \frac{50 \times 40}{0.1} \\ \\ \implies \: \frac{2000}{0.1} \\ \\ \implies \: 20000 \: N$