Question

A 0.500 kg ice puck, moving east with a speed of 3.30 m/s, has a head-on collision with a 0.900 kg puck initially at rest. Assuming a perfectly elastic collision, (a) what will be the speed of the 0.500 kg object after the collision? (b) What will be the speed of the 0.900 kg object after the collision?

Answer 1

Suppose,
The velocity of .5 kg ice puck before collision = u1= 3.30 m/s
The velocity of .9 kg ice puck before collision = u2= 0
The velocity of .5 kg ice puck after collision = v1=?
The velocity of .9 kg ice puck after collision = v2=?
According to conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
=> (u1-v1)m1 = (v2-u2)m2 ....(i)
Again, conservation of energy gives
1/2 m1u1^2+ 1/2 m2u2^2 = 1/2 m1v1^2 + 1/2 m2v2^2
=> m1(u1^2 - v1^2) = m2(v2^2 - u2^2) .....(ii)
Divide (ii) by (i)
(u1^2 - v1^2)/(u1-v1) = (v2^2 - u2^2)/(v2-u2)
=> u1+v1 = v2+u2
=> v1= u2+v2-u1 = v2-u1 ....(iii). [u2=0]
Now, substitute v1 in (i)
(u1-v2+u1)m1 = (v2-u2)m2
=> 2m1u1 - m1v2 = m2v2
=> (m1+m2)v2 = 2m1u1
=> v2 = (2m1u1)/(m1+m2) ....(iv)
Substitute v2 in (iii)
v1= v2-u1 = (2m1u1)/(m1+m2) -u1
=> v1 = (2*.5*3.30)/(.5+.9) - 3.30 = 2.357 - 3.30 = -0.943 m/s
(-) sign indicates that the ice puck is moving to the west.

Now substitute v1 in (iii) again
v1 = v2-u1
=> v2 = u1+v1 = 3.30 - 0.943 = 2.36 m/s
It means it is going to the east.

Answer 2

Correct answers.
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html#c3

Answer 3

Given:
M1 = 0.50 kg, V1 = 3.30 m/s.
M2 = 0.90 kg, V2 = 0.
V3 = Velocity of M1 after collision.
V4 = Velocity of M2 after collision.

Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.5*3.30 + 0.9*0 = 0.5*V3 + 0.9*V4,
Eq1: 0.5V3 + 0.9V4 = 1.65.

a. V3 = ((M1-M2)V1 + 2M2*V2)/(M1+M2).
V3 = (-1.32 + 0) /1.4 = -0.943 m/s. = 0.943 m/s, West.

b. In Eq1, replace V3 with -0.943 and solve for V4.
0.5 * (-0.943) + 0.9*V4 = 1.65.
V4 = 2.36 m/s, East.