Question

A 0.70 kgball moving horizontally at 5.0 m/sstrikes a vertical wall and rebounds with speed 2.0 m/s . What is the magnitude of the change in its linear momentum?

Short answer

The magnitude of the change in linear momentum of the ball is $P=4.9kg.m/s$.

Step-by-step solution

Listing the given quantities:

The mass of the ball is m = 0.70 kg .

The initial velocity of the ball is$v_i=5.0m/s$ .

The rebounding velocity of the ball is $v_f=-2.0m/s$.

Understanding the concept of law of conservation

The law of conservation of momentum states this. For two or more bodies in an isolated system acting on each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

You can use the concept of the law of conservation of momentum.

Formula:

$P=m\left(v_i-v_f\right)$

Here, m is the mass, $v_i$is the initial velocity, and $v_f$is the final velocity.

Calculations of magnitude of change in linear momentum:

The ball collides with initial velocity $v_i$and rebounds with final velocity $v_f$. Then the change in linear momentum of the ball is,

$$\begin{gathered} P=m\left(v_i-v_f\right) \\ =0.70kg\left[\left(5.0m/s\right)-\left(-2.0m/s\right)\right] \\ =4.9kg.m/s \end{gathered}$$

Hence, the change in linear momentum of the ball is $4.9kg.m/s$.