conservation of momentum

1. in the following , the white ball moves along at 2m/s to hit the black ball . the mass of the white ball and black ball are 200g and 250g respectively .

a. after collision , the black ball moves off at 0.5m/s to the right hand side . what is the final velocity of the white ball after collision

b. if the imoact lasts for 0.05s, what will be the average force on the black ball?

c.calculate the energy lost of the kinetic energy.

2. A 3kg body is fired vertically upwards from the ground with speed of 200km/h.

a.find the maximum height it can rise .

b.it's potential and kinetic energy after 1.5s

3.

Rating

1a)Let the final velocity of white ball be vw ,the inital velocity of black ball be ub. Take the inital direction of white ball tavelling positive.By conservation of momentum and energy

(2)(200/1000)+(250/1000)( ub)=(0.5)(250/1000)+( vw )(200/1000)

0.25ub=0.2vw-0.275------(1)

However the right hand side might be negative too.So

(2)(200/1000)+(250/1000)( ub)=(-0.5)(250/1000)+( vw )(200/1000)

0.25ub=0.2vw-0.525------(1")

(0.5)(0.2)(2^2)+(0.5)(0.25)(ub)^2=(0.5)(0.5)(0.25^2)+(0.5x0.2)( vw )^2

0.125(ub)^2=-0.384375+0.1(vw)^2--(2)

From (1) and (2) or (1") and (2),we get two results

I think you should upload the picture.Since the direction of the motion of white ball before impact might be towars left or right

If it is to right,use(1),to left,use (1")

1b)average force on the black ball=(change in momentum of black ball)/time taken

F=[0.25)(ub-(+/-)0.5)]/t

1c)Lost of KE=initial KE -final KE=0.5x0.2x2^2+0.5x0.25x(ub)-

(0.5x0.2x(vw)^2-(0.5)(0.25)(0.5)^2

2a)take up as positive

(0.5)(m)(v^2)=mgh

v=square root of (2gh)

(200x1000)/3600=square root of (2x10h)

h=154m

2b)By s=ut+(0.5)(a)(t^2)

s=[(200x1000)/3600]+(0.5)(-10)(1.5)^2

s=44.3m

v^2-(200x1000)/3600]^2=2(-10)(44.3)

v=46.9

so it's KE after 1.5s=0.5x3x46.9^2=3300J

PE=3x10x44.3=1329J