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

Change in momentum is
`2.667* 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* 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* 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 = (96046)/(3.6) = 2.667* 10^(4) kg.m/s`

(Since,
`1kg.km/h = (1)/(3.6) kg.m/s`)