Question

The speed of the fastest-pitched baseball was 43.0 {\rm m}/{\rm s}, and the ball's mass was 145 {\rm g}.

What was the magnitude of the momentum of this ball?
p =


Part B
How many joules of kinetic energy did this ball have?
K =




Part C -
How fast would a 58.0 gram ball have to travel to have the same amount of kinetic energy?
v =





Part D
How fast would a 58.0 gram ball have to travel to have the same amount of momentum?
v =

Answer 1

Answer with Explanation:

We are given that

Mass of ball=m=145 g=
`(145)/(1000)=0.145 kg`

1 kg=1000g

Velocity of ball=43 m/s

A.We know that

Momentum=P=mv

Using the formula

Momentum of ball,P=
`43* 0.145=6.235kg m/s`

B.Kinetic energy of ball=
`(1)/(2)mv^2=(1)/(2)(0.145)(43)^2`

Kinetic energy of ball=134.05 J

C.Mass of ball=58 g=
`(58)/(1000)=0.058 kg`

If kinetic energy=134.05 J

`k.E=(1)/(2)(0.058)v^2`

`v^2=(2* 134.05)/(0.058)=4622.41`

`v=√(4622.41)=67.99 m/s`

Velocity of ball=67.99 m/s

D.P=6.235kgm/s

`m=0.058 kg`

`P=mv`

`6.235=(0.058)v`

`v=(6.235)/(0.058)=107.5m/s`

Hence, the velocity of ball=107.5 m/s