Name: Apparent Weight in a Lift
Uploaded: 2019-07-19
Duration: 7 min 31 s
Description: In this video, we will learn about the motion of objects in an elevator, which will help you solve questions in an exam on this topic. Watch and learn.

Question

As shown in figure, two identical balls strike a rigid wall with equal speeds but at different angles of incidence. They are reflected back without any loss in speed.

(a) Determine the direction of force exerted by each ball on the wall.

(b) Determine the ratio of impulse imparted by the two balls on the wall in both cases.

Accepted Answer

(a) According to Newton's second law, the force acting on the ball is equal to change in linear momentum of the ball,

$\vec{F} = m \vec{a} = m \frac{∆ \vec{v}}{∆ t} = \frac{∆ \vec{p}}{∆ t}$ .

The direction of force is in the direction of change in linear momentum.

Case $A$ :

Change in linear momentum of the ball,

$∆ \vec{p} = \vec{p_{f}} - \vec{p_{i}} = - m v \hat{i} - m v \hat{i} = - 2 m v \hat{i}$ .

So, the force act on the ball is in the negative $x$ -direction or the direction along $X O$ .

Case $B$ :

$\vec{p_{i}} = m v \cos θ \hat{i} - m v \sin θ \hat{j}$
$\vec{p_{f}} = - m v \cos θ \hat{i} - m v \sin θ \hat{j}$
$∆ \vec{p} = \vec{p_{f}} - \vec{p_{i}} =$ $- 2 m v \cos θ \hat{i}$

So, the force is along negative $x$ -direction or along $X O$ shown in figure.

(b) Now, the calculation of impulse in both cases, respectively:

The momentum of the ball before reflection, ${\vec{p}}_{i} = m v \hat{i}$ (initial direction of motion $O X$ ),
momentum of ball after reflection, ${\vec{p}}_{f} = - m v \hat{i}$ (direction of motion after collision becomes opposite).

Now, $impulse = change in momentum$
$\vec{J_{A}} = {\vec{p}}_{f} - {\vec{p}}_{i} = - m v \hat{i} - m v \hat{i} = - 2 m v \hat{i}$ .

Hence, impulse on wall will be equal to $2 m v$ in negative $x$ -direction or along $X O$ .

Also,

$\vec{p_{i}}$ $= m v \cos θ \hat{i} - m v \sin θ \hat{j}$
$\vec{p_{f}} = - m v \cos θ \hat{i} - m v \sin θ \hat{j}$
${\vec{J}}_{B} = \vec{p_{f}} - \vec{p_{i}} = - 2 m v \cos θ \hat{i}$ .

Hence, impulse in this case is $2 m v \cos θ$ in negative $x$ -direction or along $X O$ and the ratio $= \frac{J_{A}}{J_{B}} = \frac{2 m v}{2 m v \cos θ} = \frac{1}{\cos θ}$ .

Important Questions on Newton's Laws of Motion (Without Friction)

Learn and Prepare for any exam you want !