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

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.

Step-by-step 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* 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=(J_x)/(\triangle t)` ⇒
`(F_y)_a_v_g=(J_y)/(\triangle t)`

⇒
`(F_x)_a_v_g=(0.72\ kg.ms^-^1)/(1.85* 10^-^3\ s)` ⇒
`(F_y)_a_v_g=(5.96\ kg.m.s^-^1)/(1.85* 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.