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.

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, = 0.145 kg

Velocity before the impact, = 40 m/s

Velocity after the impact, = 52 m/s

Angle of deflection, = 30 deg

Time of contact, = 1.85 ms

Conversion of milliseconds to seconds.

⇒

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

⇒

⇒ and  

⇒ Taking the vector components of 'v2'

⇒ and

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

Momentum with 'v1'

⇒

⇒

⇒

⇒  and

Momentum with 'v2'

⇒                       ⇒

⇒            ⇒

⇒         ⇒

Now the difference in terms of impulse (J).

Horizontally                         Vertically

⇒                ⇒          

⇒                 ⇒

⇒          ⇒  

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

So,

Horizontal force :                  Vertical force:

⇒                          ⇒

⇒           ⇒

⇒               ⇒

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

i think it would be 5