Question

A toy car of mass $ 2.0\,kg $ is moving towards negative $ Y $ -axis with a velocity of $ 0.5\,m/s $ . It takes a turn towards $ X $ -axis with a velocity of $ 0.4 \,m/s $ . How much is the change in the linear momentum of the car due to the turn ?

• A. $ \left(0.6\,\hat{i}-1.0\,\hat{j}\right) kg.m/s $
• B. $ \left(0.8\,\hat{i}+1.0\,\hat{j}\right) kg.m/s $
• C. $ \left(0.8\,\hat{i}-1.0\,\hat{j}\right) kg.m/s $
• D. $ \left(0.8\,\hat{i}-0.6\,\hat{j}\right) kg.m/s $

Answer 1

Answer: (B)

Given, mass of car, $m = 2\, kg$
Initial velocity, $v_i = - 0.5 \hat{j}\, m/s$
So, initial momentum $p_i = 2 \times (- 0.5 \hat{j})$
$= -1 .0 \hat{j}\, kg \,m/s$
Now, final velocity, $v_f = 0.4\hat{i} \,m/s $
So, final momentum, $p_f= 2 \times (0.4 \hat{i})$
$= 0.8 \hat{i}\, kg\, m/s$
Now, change in momentum $= p_f - p_i$
$= (0.8\hat{i}) - (n -10 \hat{j})= 0.8\hat{i} + \hat{j}$
$= (0.8 \hat{i} + 1.0\hat{ j})\, kg \,m/s$