A skater stands still on smooth ice, hands on a loaded sled. One shove sends the sled gliding one way, and the skater glides off the other way. Nobody pushed the skater, or so it seems. The numbers say how fast she goes anyway.
Before the shove, nothing moves: the system's total momentum is zero. Conservation keeps it zero, so the two momenta afterward must cancel. Two objects pushed apart from rest carry equal and opposite momenta.
Here is the scene, taking rightward as positive: a 60 kg skater and a 40 kg loaded sled, both at rest. Total momentum before: 0 kg·m/s.
The shove sends the sled right at 3 m/s. The sled's momentum after is 40 × 3 = 120 kg·m/s.
The total must still be zero. So the skater's momentum after is −120 kg·m/s.
Divide by the skater's mass: v' = −120 ÷ 60 = −2, so the skater glides left at 2 m/s.
The set-up in symbols starts from zero: 0 = m_1 v_1' + m_2 v_2'. One momentum is the other's negative.
The lighter object always leaves faster: the same size of momentum, spread over less mass, means more velocity.
The backward motion is called the recoil. The velocity the solve returns for the recoiling object is called the recoil velocity.
Take rightward as positive. A 1.5 kg cart and a 0.5 kg cart sit at rest with a compressed spring between them. The spring is released, and the 1.5 kg cart moves right at 1 m/s. Calculate the 0.5 kg cart's velocity.
Write down the values in the question:
m_1 = 1.5 kg, v_1' = +1 m/s
m_2 = 0.5 kg, v_2' = ?
Total momentum before = 0 kg·m/s (both at rest)
Write down the equation:
0 = m_1 v_1' + m_2 v_2'
Substitute in the values:
0 = 1.5 × 1 + 0.5 v_2'
0 = 1.5 + 0.5 v_2'
Simplify:
0.5 v_2' = −1.5
v_2' = −1.5 ÷ 0.5 = −3
v_2' = −3 m/s
The 0.5 kg cart moves at 3 m/s in the negative (leftward) direction.
Take rightward as positive. A 50 kg skater at rest on smooth ice throws a 2 kg ball to the right at 8 m/s. Calculate the skater's velocity just after the throw.
Write down the values in the question:
m_1 = 2 kg, v_1' = +8 m/s (the ball)
m_2 = 50 kg, v_2' = ? (the skater)
Total momentum before = 0 kg·m/s
Write down the equation:
0 = m_1 v_1' + m_2 v_2'
Substitute in the values:
0 = 2 × 8 + 50 v_2'
0 = 16 + 50 v_2'
Simplify:
50 v_2' = −16
v_2' = −16 ÷ 50 = −0.32
v_2' = −0.32 m/s
The skater drifts at 0.32 m/s in the negative (leftward) direction.
Take rightward as positive. Two roller skaters stand at rest and push apart. The 48 kg skater moves right at 2.1 m/s. Enter the 72 kg skater's velocity in m/s (include the sign).
Take rightward as positive. A 2 kg cart and a 3 kg cart sit at rest with a compressed spring between them. On release the 2 kg cart moves right at 4.2 m/s. Enter the 3 kg cart's velocity in m/s (include the sign).
Margit says: 'Two carts pushed apart from rest carry momenta equal in size and opposite in direction, whatever the two masses are.' Is she correct?
Take rightward as positive. A 70 kg skater at rest shoves a 35 kg sled, which slides right at 1.6 m/s. Enter the skater's velocity just after the shove in m/s (include the sign).
Take rightward as positive. A 0.4 kg cart and a 2 kg cart sit at rest with a compressed spring between them. On release the 0.4 kg cart moves left at 6 m/s. Enter the 2 kg cart's velocity in m/s (include the sign).
Two ice skaters push apart from rest. Which statement about their momenta just after is correct?