Question

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms.

Answer 1

Answer:

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms, find the horizontal and vertical components of the average force on the ball.

The average horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.

Explanation:

Given:

Mass of the baseball, $m$ = 0.145 kg

Velocity before the impact, $v_1$ = 40 m/s

Velocity after the impact, $v_2$ = 52 m/s

Angle of deflection, $\theta$ = 30 deg

Time of contact, $\triangle t$ = 1.85 ms

Conversion of milliseconds to seconds.

⇒ $\triangle t = 1.85\times 10^-^3\ sec$

Let the horizontal and vertical component of 'v1' and 'v2' be:

⇒ $v_1x,v_1y,v_2x,v_2y$

⇒ $v_1x=40\ ms^-^1$ and  $v_2y= 0\ ms^-^1$

⇒ Taking the vector components of 'v2'

⇒ $v_2x=52cos(30) = 45.03\ ms^-^1$ and $v_2y=52sin(30)=40.97\ ms^-^1$

We have to find the impulse in x-y, direction.

Momentum with 'v1'

⇒ $P_1_x=mv_1_x$

⇒ $P_1_x=0.145(40)$

⇒ $P_1_x=0.145(40)$

⇒ $P_1_x=5.8\ kg.m.s^-^1$  and $P_1_y=0$

Momentum with 'v2'

⇒ $P_2_x=mv_2_x$                       ⇒ $P_2_y=mv_2_y$

⇒ $P_2_x=0.145(45.03)$            ⇒ $P_2_y=0.145(40.97)$

⇒ $P_2_x=6.52\ kg.m.s^-^1$         ⇒ $P_2_y=5.96\ kg.m.s^-^1$

Now the difference in terms of impulse (J).

Horizontally                         Vertically

⇒ $J_x=P_2_x-P_1_x$                ⇒ $J_y=P_2_y-P_1_y$          

⇒ $J_x=6.52-5.8$                 ⇒ $J_y=5.96-0$

⇒ $J_x=0.72\ kg.m.s^-^1$          ⇒ $J_y=5.96\ kg.m.s^-^1$  

Now we know that Impulse is the product of  force and time.

So,

Horizontal force :                  Vertical force:

⇒  $(F_x)_a_v_g=\frac{J_x}{\triangle t}$                        ⇒ $(F_y)_a_v_g=\frac{J_y}{\triangle t}$

⇒ $(F_x)_a_v_g=\frac{0.72\ kg.ms^-^1}{1.85\times 10^-^3\ s}$           ⇒ $(F_y)_a_v_g=\frac{5.96\ kg.m.s^-^1}{1.85\times 10^-^3\ s}$

⇒ $(F_x)_a_v_g=389.18\ N$               ⇒ $(F_x)_a_v_g=3221.62\ N$

So the horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.