Question

A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in the car's momentum?

Answer 1

Answer:

Change in momentum is $2.667\times 10^{4} kg.m/s$

Solution:

The momentum of any body is the product of its mass and the velocity associated with the body and is generally given by:

$\vec{p} = m\vec{v}$

Now, as per the question:

Mass of the car, M = 1500 kg

The velocity in the east direction, $v\hat{i} = 40\hat{i} km/h$

The velocity in the north direction, $v\hat{j} = 50\hat{j} km/h$

Now, the momentum of the car in the east direction:

$p\hat{i} = mv\hat{i} = 1500\times 40\hat{i} = 60000\hat{i} kg.km/h$

Now, the momentum of the car in the north direction:

$p\hat{j} = mv\hat{j} = 1500\times 50\hat{j} = 75000\hat{j} kg.km/h$

Change in momentum is given by:

$\Delta p = p\hat{i} - p\hat{j} = 60000\hat{i} - 75000\hat{j}$

Now,

$|\Delta p| = |60000\hat{i} - 75000\hat{j}|$

$|\Delta p| = \sqrt{60000^{2} + 75000^{2}}$

$|\Delta p| = 96046 kg.km/hr = \frac{96046}{3.6} = 2.667\times 10^{4} kg.m/s$

(Since, $1kg.km/h = \frac{1}{3.6} kg.m/s$)