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?

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.

Correct answers.

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

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.

To solve this problem, we can use the principles of conservation of momentum and kinetic energy.

Firstly, let's calculate the initial momentum of each object:

The momentum (p) of an object is given by the product of its mass (m) and velocity (v):

p = m * v

For the 0.500 kg ice puck moving east with a speed of 3.30 m/s,
its initial momentum (p1) is:
p1 = 0.500 kg * 3.30 m/s = 1.65 kg·m/s

For the 0.900 kg puck initially at rest, its initial momentum (p2) is zero since its velocity is zero.

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

p_total_before = p1 + p2 = 1.65 kg·m/s

(a) To find the velocity of the 0.500 kg object after the collision, we can set up an equation using the conservation of momentum principle:

p_total_after = p1_after + p2_after

Since the first object rebounds after the collision and the second object was initially at rest, the equation becomes:

p_total_after = -p1_after + p2_after

where p1_after represents the momentum of the 0.500 kg object after the collision and p2_after represents the momentum of the 0.900 kg object after the collision.

As the collision is assumed to be perfectly elastic, kinetic energy is also conserved. The kinetic energy (K) of an object is given by the equation:

K = (1/2) * m * v^2

(b) To find the velocity of the 0.900 kg object after the collision, we can set up an equation using the conservation of kinetic energy principle:

K_total_before = K1_before + K2_before = K1_after + K2_after

Since the first object rebounds after the collision and the second object was initially at rest, the equation becomes:

K_total_before = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2

where v1 and v2 are the initial velocities of the objects, and v1' and v2' are the velocities after the collision.

By using these two equations, we can solve for both v1' and v2' and find the answer to the problem.