# Mechanics problem: momentum conservation ?

I found the first answer to be 1.04 m/s which is correct.

Then, when I tried getting the 2nd answer I got the same result which isn't correct.

Second answer with explanation please?

### 2 Answers

- WhomeLv 73 weeks ago
A couple of items come to mind.

1) A limit of two significant digits are apparent in the question numerals. Answers should be limited to the same precision.

2) The question states that the velocity is relative to the self, not the ground. So the impulse is possibly less than obvious.

Conservation of momentum both weights at the same time.

Let V be the velocity of the thrown weights (assumed negative) relative to ground. Let v be the velocity of rider and cart (assumed positive) after throw. |V| + |v| = 2.5 m/s.

To conserve momentum, the velocity of each mass relative to ground will be reversely proportional to ratio of the mass departing to the original whole mass.

The velocity of the weights will be 2.5 m/s((45 + 15) / (45 + 15 + 2(12.5)) = 1.764 m/s

The velocity of the cart and rider will be 2.5 m/s (2(12.5)) / (45 + 15 + 2(12.5)) = 0.7352 m/s.

We can see that the sum of these two velocities is within rounding errors to 2.5 m/s.

So rounding to 2 s.d. the cart velocity is 0.74 m/s.

Conservation of momentum, weights one at a time.

cart velocity, first weight, 2.5 m/s(12.5) / (45 + 15 + 2(12.5)) = 0.3676 m/s.

Now if we allow the reference frame to move with this velocity, we can see the affect of the second weight 2.5(12.5) / (45 + 15 + 12.5) = 0.4310 m/s.

So the total velocity of the cart will be the sum of these two. 0.3676 + 0.4310 = 0.7987

With rounding to 2 s.d. the cart velocity is 0.80 m/s.

I feel a little bit sorry for you and your class as this seems to me to be either a very poorly posed question or an incorrect solution to the question as posed

.