# Physics Help please?

A sandbag is mounted on a cart that is at rest on a frictionless surface, and their total mass is 4.5 kg. What will be the velocity of the cart and sandbag if a bullet of a mass of 2.0 g is fired into the sandbag with a horizontal velocity of 500 m/s?

### 5 Answers

- 3 weeks agoFavorite Answer
4.5 kg = 4500 g

m1 * v1 + m2 * v2 = m3 * v3

m3 = m1 + m2 = 4500 + 2 = 4502 grams

4500 * 0 + 2 * 500 = 4502 * v

1000 = 4502 * v

1000 / 4502 = v

v = 0.22212350066637050199911150599733...

0.222 m/s

- oubaasLv 73 weeks ago
Momentum shall be conserved, therefore :

V = (mb*Vb)/(mb+mc) = 2*500/(1000*4.502) = 1/4.502 m/sec ( 0.2221... )

lost energy LE = (5^2*10^4*10^-3)-4.502/2*(1/4.502)^2 = 250-0.11 = 249.9 joule

- 3 weeks ago
Since momentum is conserved, and it is an inelastic collision since they stick together;

m1v1 + m2v2 = (m1 + m2)v3, where v3 is the final velocity of the cart and sandbag

v2 = 0 since the cart is at rest, and m2 = 0.002 kg, therefore

v3 = (m1v1)/(m1+m2) = 0.222m/s

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- Andrew SmithLv 73 weeks ago
convert the mass to kg 2 g -> 0.002 kg which is smaller than the imprecision in total mass.

MOMENTUM IS CONSERVED 0.002 * 500 = 4.5 (+0.002) * V

v= 0.002*500/4.5 ~= 0.22 m/s

One thing you should consider. Both "4.5" and "2.0" contained only 2 significant figures so even 0.222 is claiming more accuracy than the question permitted.