Question

A 1450 kg car moving south at 11.1 m/s collides with a 2460 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.01 m/s to the north. Find the velocity of the 2460 kg car before the collision.

Answer 1

14.51m/s north

Step-by-step explanation:

Let the first car be body A

Let the second car be body B

Given two bodies A and B undergoing collision, using the principle of conservation of linear momentum that total momentum (
`P_(A1)` +
`P_(B1)`) of two bodies before collision is equal to the total momentum (
`P_(A2)` +
`P_(B2)`) of the bodies after collision. i.e

`P_(A1)` +
`P_(B1)` =
`P_(A2)` +
`P_(B2)` ---------------(i)

Where;

`P_(A1)` = Momentum of body A before collision

`P_(B1)` = Momentum of body B before collision

`P_(A2)` = Momentum of body A after collision

`P_(B2)` = Momentum of body B after collision

Also;

`P_(A1)` =
`m_(A)` x
`u_(A)` [
`m_(A)` = mass of body A,
`u_(A)` = initial velocity of body A]

`P_(B1)` =
`m_(B)` x
`u_(B)` [
`m_(B)` = mass of body B,
`u_(B)` = initial velocity of body B]

`P_(A2)` =
`m_(A)` x
`v_(A)` [
`m_(A)` = mass of body A,
`u_(A)` = final velocity of body A]

`P_(B2)` =
`m_(B)` x
`v_(B)` [
`m_(B)` = mass of body B,
`v_(B)` = final velocity of body B]

Substitute these values into equation (i) as follows;

(
`m_(A)` x
`u_(A)`) + (
`m_(B)` x
`u_(B)`) = (
`m_(A)` x
`v_(A)`) + (
`m_(B)` x
`v_(B)`) ------------------(ii)

But since the two move together after collision, then their final velocities are the same. i.e
`v_(A)` =
`v_(B)` = v

Therefore, equation (ii) becomes;

=> (
`m_(A)` x
`u_(A)`) + (
`m_(B)` x
`u_(B)`) = (
`m_(A)` x v) + (
`m_(B)` x v)

=> (
`m_(A)` x
`u_(A)`) + (
`m_(B)` x
`u_(B)`) = v (
`m_(A)` +
`m_(B)` ) ----------------------------(iii)

Let south direction be negative and north direction be positive;

Therefore, from question;

`m_(A)` = 1450kg

`u_(A)` = - 11.1m/s [south is negative]

`m_(B)` = 2460kg

`u_(B)` = ?

v = +5.01m/s [north is positive]

Substitute these values into equation (iii) as follows;

=> (1450 x -11.1) + (2460 x
`u_(B)`) = 5.01(1450 + 2460)

=> (-16095) + (2460
`u_(B)`) = 19589.1

=> -16095 + 2460
`u_(B)` = 19589.1

=> 2460
`u_(B)` = 19589.1 + 16095

=> 2460
`u_(B)` = 35684.1

=>
`u_(B)` = 35684.1 / 2460

=>
`u_(B)` = 14.51 m/s

Therefore, the velocity of body B (second car with 2460kg) before collision is 14.51m/s and since it is positive it shows that it is in the north direction which confirms the situation in the question.