Two carts meet head-on and bounce apart. A motion sensor catches one cart's rebound, but the other cart left the sensor's range. Conservation of momentum fills in the missing velocity.
In a bounce-apart collision the set-up keeps two after-velocities: m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'. When three of the four velocities are measured, the equation hands you the fourth.
Here is the scene, taking rightward as positive: a 2 kg cart at +5 m/s meets a 4 kg cart at −1 m/s, and they bounce apart. The sensor catches the 2 kg cart rebounding at −1 m/s.
Total the before side: 2 × 5 + 4 × (−1) = 6, so the system carries 6 kg·m/s.
Write the after side with the measured rebound in place: 2 × (−1) + 4 v_2' = 6.
The 2 kg cart's after-momentum is −2 kg·m/s. So 4 v_2' = 6 − (−2) = 8.
Divide: v_2' = 8 ÷ 4 = 2, so the 4 kg cart leaves at +2 m/s, rightward.
The signs do the direction work. A rebound enters as a negative value, and subtracting a negative adds size to what remains.
Take rightward as positive. A 3 kg cart at +2 m/s meets a 1 kg cart at −2 m/s head-on, and they bounce apart. Afterward the 3 kg cart moves at +0.5 m/s. Calculate the 1 kg cart's velocity after the collision.
Write down the values in the question:
m_1 = 3 kg, v_1 = +2 m/s, v_1' = +0.5 m/s
m_2 = 1 kg, v_2 = −2 m/s, v_2' = ?
Write down the equation:
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
Substitute in the values:
3 × 2 + 1 × (−2) = 3 × 0.5 + 1 v_2'
6 − 2 = 1.5 + v_2'
4 = 1.5 + v_2'
Simplify:
v_2' = 4 − 1.5 = 2.5
v_2' = +2.5 m/s
The 1 kg cart leaves at 2.5 m/s in the positive (rightward) direction.
Take rightward as positive. A 2 kg cart at +3 m/s meets a 2 kg cart at −1 m/s, and they bounce apart. Afterward the first cart is at rest. Calculate the second cart's velocity after the collision.
Write down the values in the question:
m_1 = 2 kg, v_1 = +3 m/s, v_1' = 0 m/s
m_2 = 2 kg, v_2 = −1 m/s, v_2' = ?
Write down the equation:
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
Substitute in the values:
2 × 3 + 2 × (−1) = 2 × 0 + 2 v_2'
6 − 2 = 2 v_2'
Simplify:
v_2' = 4 ÷ 2 = 2
v_2' = +2 m/s
The second cart leaves at 2 m/s in the positive (rightward) direction.
Take rightward as positive. A 2 kg cart at +6 m/s meets a 4 kg cart at −2 m/s, and they bounce apart. Afterward the 2 kg cart moves at −1.4 m/s. Enter the 4 kg cart's velocity after the collision in m/s (include the sign).
Take rightward as positive. A 1 kg puck at +4 m/s hits a parked 2 kg puck, and the pucks bounce apart. Afterward the 1 kg puck moves at −0.5 m/s. Enter the 2 kg puck's velocity after the collision in m/s (include the sign).
Take rightward as positive. A 3 kg cart at +4 m/s meets a 2 kg cart at −3 m/s, and they bounce apart. Afterward the 2 kg cart moves at +2.4 m/s. Enter the 3 kg cart's velocity after the collision in m/s (include the sign).
Take rightward as positive. A 4 kg cart at +2.5 m/s meets a 1 kg cart at −4 m/s, and they bounce apart. Afterward the 4 kg cart moves at +0.2 m/s. Enter the 1 kg cart's velocity after the collision in m/s (include the sign).
Take rightward as positive. A 6 kg cart at +1 m/s meets a 2 kg cart at −6 m/s, and they bounce apart. Afterward the 2 kg cart moves at +1.8 m/s. Enter the 6 kg cart's velocity after the collision in m/s (include the sign).
Two carts bounce apart in a one-dimensional collision, and cart 2's final velocity is the one unknown. Which expression gives it?