Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 55.0 m/s, and it leaves the bat traveling to the left at an angle of 30 degrees above horizontal with a speed of 60.0 m/s. The ball and bat are in contact for 1.65 ms.

Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right. Use two significant figures

Answer 1

Fx=9399.39 iN

Step-by-step explanation:

`m_(b)=0.145 kg\\v_(b1)=55 (m)/(s)\\v_(b2)=60(m)/(s)\\\alpha=30`

Δt=1.65ms ⇒
`1.65x10^(-3) s`

Momentum of the motion is

`p=m*v`

so:

∑F= p2-p1/Δt

`p1=0.145kg*55 i\\p1=7.975 i`

`p2=0.145*(-60*cos(30)i+60*(sen(30)j)\\p2=0.145*(-51.96i+30j)\\p2=-7.534i+4.35j`

`F= ((-7.534i+4.35j)-(7.975))/(1.65x10^(-3)s)\\Fx=(15.509i)/(1.65x^(-3)s) \\Fx=9399.39i N\\Fy=(4.35j)/(1.65x^(-3)s) \\Fy=2636.36j N`