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=$\frac{145}{1000}=0.145 kg$

1 kg=1000g

Velocity of ball=43 m/s

A.We know that

Momentum=P=mv

Using the formula

Momentum of ball,P=$43\times 0.145=6.235kg m/s$

B.Kinetic energy of ball=$\frac{1}{2}mv^2=\frac{1}{2}(0.145)(43)^2$

Kinetic energy of ball=134.05 J

C.Mass of ball=58 g=$\frac{58}{1000}=0.058 kg$

If kinetic energy=134.05 J

$k.E=\frac{1}{2}(0.058)v^2$

$v^2=\frac{2\times 134.05}{0.058}=4622.41$

$v=\sqrt{4622.41}=67.99 m/s$

Velocity of ball=67.99 m/s

D.P=6.235kgm/s

$m=0.058 kg$

$P=mv$

$6.235=(0.058)v$

$v=\frac{6.235}{0.058}=107.5m/s$

Hence, the velocity of ball=107.5 m/s