Physics, 20.02.2020 19:30 mikey952

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.

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Answer from: jaanne

14.51m/s north

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.

Answer from: Quest

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