Lee is rolling along her 4 kg skateboard with a constant speed of 3 m/s when she jumps off the back and continues forward with a velocity of 2 m/s relative to the ground. This causes the skateboard to go flying forward with a speed of 15.5 m/s relative to the ground. What is Lee's mass?
2007-11-17 16:13:11 · 3 answers · asked by Light Up the Sky 1 in Science & Mathematics ➔ Physics
Let M be the mass of Lee. The initial momentum is
(M + 4 kg)(3 m/s)
Lee jumps off. Momentum is conserved. The skateboard has momentum (4 kg)(15.5 m/s) and Lee has momentum M(2 m/s)
(M + 4)(3) = (4)(15.5) + M(2)
Solve for M
2007-11-17 16:18:39 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
Use conservation of momentum:
(x+4 kg)(3 m/s) = (x)(2 m/s) + (4 kg)(15.5 m/s)
Solve for x
3x + 12 = 2x + 64
x = 52 kg
2007-11-17 16:21:09 · answer #2 · answered by BurningPyre 4 · 0⤊ 0⤋
This is a basic conservation of momentum problem. The momentum of the person/skateboard system before lee jumps off is equal to it's momentum after.
Call lee's mass (x)
momentum = mass times velocity
(4kg + x)(3m/s) = (x)(2m/s) + (4kg)(15.5m/s)
Then just solve for (x) and you have your answer
2007-11-17 16:20:25 · answer #3 · answered by Aaron777 3 · 0⤊ 0⤋