Question

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a momentum of magnitude 4.0 kg∙ m/s. What is the magnitude of the large object's momentum change?

Answer 1

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$, $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$, $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$, $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$, then the final momentum of the big object is:

$7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}$

$p_{B,2} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is:

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}$

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is 3 kilogram-meters per second.