Question

A bat strikes a 0.145 kgkg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/sm/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘30∘ above horizontal with a speed of 38.0 m/sm/s. The ball and bat are in contact for 1.75 msms. find the horizontal and vertical components of the average force on the ball.

Answer 1

Answer with Explanation:

We are given that

Mass of ball=m=0.145 kg

Initially horizontal  velocity of ball,ux=50 m/s

$\theta=30^{\circ}$

Final velocity of ball,v=38m/s

Time ,t=1.75 ms=$1.75\times 10^{-3} s$

$1 ms=10^{-3} s$

Horizontal component of average force, $F_x=\frac{m(vcos\theta-u_x)}{t}$

Using the formula

Horizontal component of average force, $F_x=\frac{0.145(-38cos30-50)}{1.75\times 10^{-3}}=-6.9\times 10^3$N

Vertical component of average force, $F_y=\frac{m(vsin\theta-u_y)}{t}$

Vertical component of average force,$F_y=\frac{0.145(38sin30-0}{1.75\times 10^{-3}}=1.6\times 10^3 N$