# Momentum textbook problem

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1. Feb 22, 2015

### x86

1. The problem statement, all variables and given/known data
The 75kg boy leaps off the cart A with a horizontal velocity of v' = 3 m/s to the left measured relative to the cart. Determine the velocity of cart A just after the jump. If he then lands on cart B with the same velocity that he left cart A, determine the velocity of cart B just after he lands on it. Carts A and B have the same mass of 50kg and are originally at rest.

2. Relevant equations
Relative motion: Va = Va/b + Vb
Momentum: m1V1 = m2V2

3. The attempt at a solution
the subscript j will represent the boy, while a/b represent the respective carts. the coordinate system: left represents negative, right positive

Relative motion
Vj = Vj/a + Va
(1) Vj = -3 + Va

Momentum
0 + 0 = ma*Va - mj*Vj
mj*Vj = ma*Va
(2) Vj = (ma)/(mj)Va = (50/75)Va

Substituting (1) and (2)
(50/75)Va = -3 + Va
Va(50/75-1)=-3
Va = 9 m/s (to the right)
Vj = -3 + 9m/s = 6m/s (to the left)

Now I find Vb (assume direction right)
mj*Vj + mb*Vb = (mj+mb)V
75*(-6) + 50*0 = (50+75)V

V = (75*6)/(50+75) = 3.6 (left)

But this is wrong,the answer is 0.720 m/s (left)

2. Feb 22, 2015

### haruspex

Need to be consistent about signs. Are you measuring all velocities as positive right, or the cart's as positive right and the boy's as positive left?
What relative velocity would that be?