Asked by Pat
A 44 kg child jumps off a 2.2 kg skateboard that was moving at 8.0 m/s. The skateboard comes to a stop as a result. Find the speed at which the child jumped from the board. Show all your work. Assume a frictionless, closed system.
Answers
Answered by
Damon
original momentum = 46.2 * 8
Final momentum = original momentum
= 44 * v - 2.2 * 0
so
46.2 * 8 = 44 * v
Final momentum = original momentum
= 44 * v - 2.2 * 0
so
46.2 * 8 = 44 * v
Answered by
drwls
Apply the law of conservation of momentum.
The child must have jumped on while traveling in the opposite direction
m*8.0 m/s + M*Vchild = (M + m) *Vfinal = 0
m is the skateboard mass and M is the child's mass.
Solve for Vchild
Vchild = -(m/M)*8.0 m/s
The child must have jumped on while traveling in the opposite direction
m*8.0 m/s + M*Vchild = (M + m) *Vfinal = 0
m is the skateboard mass and M is the child's mass.
Solve for Vchild
Vchild = -(m/M)*8.0 m/s
Answered by
Damon
The child jumped off :)
Answered by
drwls
whoops
Answered by
kage
i needz hlep
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