Calculate Impulse from Baseball Bat Strike - 20 kgm/s

[logglypop]

A miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.500 m/s. What is the bullet's speed?
Wnet= 1/2mv^2-1/2mv^2
0=1/2mv^2-1/2mv^2
1/2mv^2=1/2mv^2
1/2(.oo5)v^2=1/2(.120)(.5)^2
v= 2.5 m/s
i don't think the answer reasonable, but that the best i can come up with

A baseball player *****es a .20kg baseball. The ball arrive the home plate at 40m/s and is batted straight back with speed 60m/s. The bat contact with the ball for .05s, find impulse
P=mv final - mv initial
p= .2(40)- .2( -60)
p=20kgm/s

i need advice please

 

[mgb_phys]

A miniature spring-loaded, radio-controlled gun is mounted on an air puck.

Canada's secret weapons program!

v= 2.5 m/s
i don't think the answer reasonable, but that the best i can come up with

That's what I get - note you don't have to convert the masses into kg since the units cancel.

A baseball player *****es a .20kg baseball. The ball arrive the home plate at 40m/s and is batted straight back with speed 60m/s. The bat contact with the ball for .05s, find impulse
P=mv final - mv initial
p= .2(40)- .2( -60)
p=20kgm/s

Impulse is just change in momentum so that looks right too.
Hint Impulse = force * time, and f=ma so impulse = m a t = Kg *m/s^2 *s = kgm/s so your units are right