A bullet is far too fast for a stopwatch. Fired into a block on a low-friction track, though, it leaves a slow, measurable glide. Conservation of momentum runs the collision backward and returns the bullet's speed.
Every solve so far ran forward: before-values in, after-values out. The same equation runs the other way. When the outcome is measured, the one unknown can sit on the before side.
Here is the scene, taking rightward as positive: a 0.010 kg bullet is fired into a 1.990 kg block, parked on the track. The bullet embeds, and the pair glides off at 2 m/s.
The after side is fully known: the combined 2.000 kg moves at 2 m/s, so the total momentum is 2.000 × 2 = 4 kg·m/s.
Conservation says the before total was also 4 kg·m/s. Before the shot, only the bullet moved.
So the bullet carried all 4 kg·m/s on its 0.010 kg: v = 4 ÷ 0.010 = 400, so the bullet flew at 400 m/s.
The tiny mass is why the pair is so slow: the same momentum spread over 200 times the mass gives one two-hundredth of the velocity.
The target does not have to be parked. If both objects move before the collision, the known object's before-momentum joins the bookkeeping, and the unknown velocity still falls out of one equation.
Take rightward as positive. A 0.05 kg dart is fired into a parked 0.45 kg foam cart. The dart embeds, and the pair rolls off at 1.2 m/s. Calculate the dart's speed before the hit.
Write down the values in the question:
m_1 = 0.05 kg, v_1 = ?
m_2 = 0.45 kg, v_2 = 0 m/s
shared v' = +1.2 m/s
Write down the equation:
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'
Substitute in the values:
0.05 v_1 + 0 = 0.5 × 1.2
0.05 v_1 = 0.6
Simplify:
v_1 = 0.6 ÷ 0.05 = 12
v_1 = +12 m/s
The dart flew at 12 m/s before the hit.
Take rightward as positive. A 2 kg cart rolls right at an unknown speed and catches a 1 kg cart rolling right at 1 m/s. They couple and move at 2 m/s. Calculate the 2 kg cart's speed before the collision.
Write down the values in the question:
m_1 = 2 kg, v_1 = ?
m_2 = 1 kg, v_2 = +1 m/s
shared v' = +2 m/s
Write down the equation:
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'
Substitute in the values:
2 v_1 + 1 × (+1) = 3 × 2
2 v_1 + 1 = 6
Simplify:
2 v_1 = 6 − 1 = 5
v_1 = 5 ÷ 2 = 2.5
v_1 = +2.5 m/s
The 2 kg cart rolled at 2.5 m/s before the collision.
Take rightward as positive. A 0.2 kg arrow is fired into a parked 3.8 kg target cart. The arrow embeds, and the pair rolls off at 0.8 m/s. Enter the arrow's speed before the hit in m/s.
Take rightward as positive. A 3 kg cart rolls right at an unknown speed and catches a 2 kg cart rolling right at 1.2 m/s. They couple and move at 3 m/s. Enter the 3 kg cart's speed before the collision in m/s.
Take rightward as positive. A 1 kg cart moving right at an unknown speed meets a 3 kg cart moving left at 2 m/s head-on. They couple and move at +0.5 m/s. Enter the 1 kg cart's speed before the collision in m/s.
Take rightward as positive. A 0.3 kg snowball hits the back of a parked 2.7 kg sled and stays pressed onto it. The pair slides off at 1.1 m/s. Enter the snowball's speed before the hit in m/s.
Take rightward as positive. A 1200 kg car moving right at an unknown speed rear-ends an 800 kg car moving right at 9 m/s. The bumpers lock, and the pair slides off at 12 m/s. Enter the 1200 kg car's speed before the impact in m/s.
An arrow embeds in a parked cart, and the pair rolls off far slower than the arrow flew. Why is the pair's speed so much smaller?