Conservation of momentum problem from SAT II

[incog13]

Hello,

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.

 

[nasu]

Momentum is a vector so use vector addition, considering that "He throws the box behind him".

 

[bigerst]

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

.

which direction do u define to be positive? remember momentum is a VECTOR quantity.

 

[incog13]

Ok, so let me rethink this. If the initial direction of motion of man + box is set to be positive, then am I simply going to be adding two positive quantities in the initial calculation of the momentum?

If the final velocity of the box is 5 m / s with respect to the ground, and I am calculating the final momentum also with respect to the ground, then I am guessing that the final calculation is also going to be adding two positive velocities? Or no?

I am kind of confused .

 

[jahaan]

I think the problem should be interpreted as: "he throws the box, so it travels away from him at 5 m/s". Then use galileo's principle of relativity (to find the speed of the box with respect to the ground), and calculate the momentum.

 

[jbriggs444]

Ok, so let me rethink this. If the initial direction of motion of man + box is set to be positive, then am I simply going to be adding two positive quantities in the initial calculation of the momentum?

If the final velocity of the box is 5 m / s with respect to the ground, and I am calculating the final momentum also with respect to the ground, then I am guessing that the final calculation is also going to be adding two positive velocities? Or no?

I am kind of confused .

You are not confused. You are correct. You solved the problem correctly. Kaplan posed it incorrectly.

When the man throws the box behind him there are two velocities which would give the box a "speed" of 5 meters/sec with respect to the ground.

1. He throws it backwards so that it attains a velocity of -5 meters/sec with respect to the ground.

2. He throws it backwards so that it attains a velocity of +5 meters/sec with respect to the ground.

These are the only interpretations that make the phrase "with respect to the ground" meangful. Since the man's "velocity" is referred to with a sign convention where positive is forward, it is reasonable to assume that the resulting velocity of the box is also positive.

Accordingly:

Interpretation 1 is a correct interpretation.
Interpretation 2 is the intended interpretation and yields the 11 meters/sec result.

 

[incog13]

Thank you jbriggs444! I am not going out of my mind after all. Phew!

Everyone, thanks for the replies.

 

[nasu]

You are not confused. You are correct. You solved the problem correctly. Kaplan posed it incorrectly.

When the man throws the box behind him there are two velocities which would give the box a "speed" of 5 meters/sec with respect to the ground.

1. He throws it backwards so that it attains a velocity of -5 meters/sec with respect to the ground.

2. He throws it backwards so that it attains a velocity of +5 meters/sec with respect to the ground.

These are the only interpretations that make the phrase "with respect to the ground" meangful. Since the man's "velocity" is referred to with a sign convention where positive is forward, it is reasonable to assume that the resulting velocity of the box is also positive.

Accordingly:

Interpretation 1 is a correct interpretation.
Interpretation 2 is the intended interpretation and yields the 11 meters/sec result.

Thee is no confusion. The actual signs (plus or minus) are arbitrary, depending on the choice of positive direction. What is not arbitrary is that the speed (and momentum) of the box is backwards so the sign of the box's momentum is opposite to the sign of the man's momentum, after the throw.

 

[jbriggs444]

Thee is no confusion. The actual signs (plus or minus) are arbitrary, depending on the choice of positive direction. What is not arbitrary is that the speed (and momentum) of the box is backwards so the sign of the box's momentum is opposite to the sign of the man's momentum.

We are told that the box was thrown backwards, not that it attained a backward velocity.

 

[azizlwl]

giving it a velocity of 5 m/s. with respect to the ground.

 

[nasu]

We are told that the box was thrown backwards, not that it attained a backward velocity.

Yes, you are right. I see your point.