Question

A 40 kg boy standing initially motionless on frictionless ice throws a 2.2 kg rock away from himself at a speed of 5.1 m/s. What is the boy's speed after he throws the rock?

a) 0.81 m/s
b) 1.62 m/s
c) 2.43 m/s
d) 3.24 m/s

Answer 1

The boy's speed after throwing the rock is 0.81 m/s in the opposite direction.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The initial momentum of the boy and the rock is zero, since they are both initially motionless. After the boy throws the rock, he will experience an equal and opposite momentum in the backward direction. This momentum will cause the boy to move in the opposite direction.

We can calculate the boy's speed after throwing the rock using the equation:

mboyvboy + mrockvrock = 0

Where:

mboy = mass of the boy = 40 kg

vboy = final velocity of the boy (unknown)

mrock = mass of the rock = 2.2 kg

vrock = velocity of the rock after being thrown = 5.1 m/s

Substituting the values into the equation, we get:

40 kg * vboy + 2.2 kg * 5.1 m/s = 0

Solving for vboy, we find that the boy's speed after throwing the rock is -0.81 m/s.

Since speed is a scalar quantity and direction is indicated by the positive/negative sign, the boy's speed is 0.81 m/s in the opposite direction of the rock's velocity.