Question

A 30-kg girl and a 25-kg boy are standing on frictionless roller skates. The girl pushes the boy, who moves off at 1.0 m/s. The girl's speed is:

a) 0.45 m/s
b) 0.55 m/s
c) 0.83 m/s
d) 1.2 m/s

Answer 1

Using the law of conservation of momentum for a girl and boy on frictionless roller skates, the girl's speed after pushing the boy is calculated to be 0.83 m/s, which corresponds to option (c).

Step-by-step explanation:

The problem explores the conservation of momentum between a 30-kg girl and a 25-kg boy standing on frictionless roller skates. According to the law of conservation of momentum, the total momentum before the push must be equal to the total momentum after the push. Initial total momentum is zero since both are stationary. After the girl pushes the boy, the boy starts moving with a momentum of 25 kg × 1.0 m/s = 25 kg·m/s (to the right). To conserve momentum, the girl must have an equal and opposite momentum. Therefore, her velocity (vgirl) can be determined using the formula:

mgirl × vgirl = mboy × vboy

30 kg × vgirl = 25 kg × 1.0 m/s

vgirl = (25 kg × 1.0 m/s) / 30 kg

vgirl = 0.833 m/s

The girl's speed is therefore 0.83 m/s, which corresponds to option (c).