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I am having trouble understanding this short problem from a Kaplan SAT II prep book. Here it is:

A 60 kg man holding a 20 kg box rides on a skateboard at a speed of 7 m/s. He throws the box behind him, giving it a velocity of 5 m/s. with respect to the ground. What is his velocity after throwing the object?

(A) 8 m/s

(B) 9 m/s

(C) 10 m/s

(D) 11 m/s

(E) 12 m/s

This is how I thought about it: Before man threw box, momentum of man + box system was:

(7 m/s) (20 kg) + (7 m/s) (60 kg) = 560 m * kg / s

The final momentum of man + box system after box was thrown was:

(5 m/s) (20 kg) + (x m/s) (60 kg) = 560 m * kg / s

x = 7.67 m / s

So man's velocity with respect to ground after he threw the box would be 7.67 m / s, or answer A, roughly 8 m / s ?

But unfortunately, the answer guide gives me the answer of 11 m / s for the man's final speed. Please help me make some sense of this!

Thanks in advance.

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# Homework Help: Conservation of momentum problem from SAT II

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