Question

A 2 kg mass moving to the right makes an elastic head-on collision with a 4 kg mass moving to the left at 4 m/s. The 2 kg mass reverses direction after the collision and moves at 3 m/s. The 4 kg mass moves to the left at 1 m/s. What was the initial velocity of the 2 kg mass?
a
4 m/s to the left
b
1 m/s to the left
c
3 m/s to the right
d
4 m/s to the right

Answer 1

To solve this problem, we can use the principles of conservation of momentum and the fact that kinetic energy is conserved in an elastic collision.

Let the initial velocity of the 2 kg mass be \( v_1 \) (which we need to find). The initial velocity of the 4 kg mass is \( v_2 = -4 \) m/s (to the left is considered negative).

After the collision, the 2 kg mass moves at \( 3 \) m/s to the left (which is negative), and the 4 kg mass moves at \( 1 \) m/s to the left (again, negative).

### Conservation of Momentum:
The total momentum before the collision equals the total momentum after the collision.

So we set up the equation:

\[
\text{Initial momentum} = \text{Final momentum}
\]

\[
(2 \, \text{kg} \cdot v_1) + (4 \, \text{kg} \cdot -4 \, \text{m/s}) = (2 \, \text{kg} \cdot -3 \, \text{m/s}) + (4 \, \text{kg} \cdot -1 \, \text{m/s})
\]

Now, simplify the equation:

Initial momentum:

\[
2v_1 - 16
\]

Final momentum:

\[
-6 - 4 = -10
\]

Setting the two sides equal:

\[
2v_1 - 16 = -10
\]

Now, solve for \( v_1 \):

\[
2v_1 = -10 + 16
\]

\[
2v_1 = 6
\]

\[
v_1 = 3 \, \text{m/s}
\]

Since the 2 kg mass is moving to the right, \( v_1 = 3 \, \text{m/s} \).

Thus, the initial velocity of the 2 kg mass was **3 m/s to the right**. Therefore, the answer is:

**c) 3 m/s to the right**.