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 30â above horizontal with a speed of 52.0 m/s. If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Answer 1

Answer:

Fx = -7042.86N (Taking the right as positive sense of motion)

Fy = 2154.29N.

Explanation:

The full solution can be found in the attachment below. The approach to the solution involves the summation of the respective components of the momentum along the horizontal and vertical axis.

To the right is taken as positive sense of motion and the left as negative sense of motion.

The basic statement of Newtown's second law have been used which is

M(v2 – v1) = Ft

See the attachment below for the complete calculation procedure.

Answer 2

Answer:

Horizontal component, Fx =  -7045.59 N (Leftwards)

Vertical component, Fy =  2982.96 N (Upwards)

Explanation:

Mass of the baseball = 0.145 kg

Before impact, v₁ = 40 m/s

After impact, v₂ = 52 m/s

time of impact, t = 1.75 ms = 0.00175 s

According to the second law of motion, F = m(v₂ - v₁)/t

Resolving vectorially,

v₁ = 40i m/s ( right = +ve x)

v₂ = -52 cos 30 i + 52 sin 30 j m/s ( to the left (-ve x) and upwards (+ve y))

v₂ = -45.03i + 36j m/s

F = m(v₂ - v₁)/t

F = 0.145 (-45.03i + 36j - 40i)/ 0.00175

F = 0.145 (-85.03i + 36j)/ 0.00175

F = 82.86 (-85.03i + 36j)

F = -7045.59i + 2982.96j N

Fx =  -7045.59 N (Leftwards)

Fy =  2982.96 N (Upwards)