Question

Two bumper cars collide in a perfectly elastic collision. Bumper car 1 has a mass of 50 kg and an initial
velocity of 1.4 m/s. Bumper car 2 has a mass of 50 kg and an initial velocity of -2.1 m/s. After the
collision, Bumper car 1 has a velocity of -1.9 m/s. What is the velocity of bumper car 2 after the
collision?
answers:
-0.7 m/s

1.95 m/s

1.2 m/s

-1.0 m/s

Answer 1

In a perfectly elastic collision, both momentum and kinetic energy are conserved.

Let's start by applying the conservation of momentum:

\[
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
\]

where
- \(m_1\) = mass of bumper car 1 = 50 kg,
- \(u_1\) = initial velocity of bumper car 1 = 1.4 m/s,
- \(m_2\) = mass of bumper car 2 = 50 kg,
- \(u_2\) = initial velocity of bumper car 2 = -2.1 m/s,
- \(v_1\) = final velocity of bumper car 1 = -1.9 m/s,
- \(v_2\) = final velocity of bumper car 2 (unknown).

Plugging in the values into the momentum conservation equation:

\[
50 \cdot 1.4 + 50 \cdot (-2.1) = 50 \cdot (-1.9) + 50 \cdot v_2
\]

Calculating the left side:

\[
50 \cdot 1.4 = 70
\]
\[
50 \cdot (-2.1) = -105
\]
So:
\[
70 - 105 = -35
\]

Now substituting in the momentum conservation equation:

\[
-35 = 50 \cdot (-1.9) + 50 \cdot v_2
\]

Calculating \(50 \cdot (-1.9)\):

\[
50 \cdot (-1.9) = -95
\]

Now substituting that back in:

\[
-35 = -95 + 50 v_2
\]

Adding 95 to both sides:

\[
-35 + 95 = 50 v_2
\]
\[
60 = 50 v_2
\]

Solving for \(v_2\):

\[
v_2 = \frac{60}{50} = 1.2 \text{ m/s}
\]

So, the velocity of bumper car 2 after the collision is **1.2 m/s**.