Question

A 36427 kg car is traveling at a speed of 137 m/s along a straight path when it strikes a 35660.2 kg stationary car and couples to it. What is the combined speed after impact?

Answer 1

Given:

A 36427 kg car is traveling at a speed of 137 m/s along a straight path when it strikes a 35660.2 kg stationary car and couples (i.e. attaches) to it.

To find:

Combined speed after impact?

Calculation:

In this type of questions, it is best to apply CONSERVATION OF LINEAR MOMENTUM principle:

• Let final velocity of combination after impact be v.

$\rm \: m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2})v$

$\rm \implies\: (36427 \times 137) +(35660 \times 0)= (36427 + 35660)v$

$\rm \implies\: (36427 \times 137) = (72087)v$

$\rm \implies\: 4990499= (72087)v$

$\rm \implies\: v = \dfrac{4990499}{72087}$

$\rm \implies\: v = 69.2 \: m/s$

So, velocity of combination after impact is 69.2 m/s.

Answer 2

Explanation:

Given:-

$\sf m_1=36427kg \\ \sf m_2=35660kg \\ \sf u_1=137m/s \\ \sf u_2=0m/s$

To find:-

Combined speed=v

Solution:-

We know that

$\boxed{\sf m_1u_1+m_2u_2=(m_1+m_2)v}$

$\\ \tt{:}\dashrightarrow (36427\times 137)+(35660\times0)=(36427+35660)v$

$\\ \tt{:}\dashrightarrow 4990499+0=72087v$

$\\ \tt{:}\dashrightarrow 4990499=72087v$

$\\ \tt{:}\dashrightarrow v=\dfrac{4990499}{72087}$

$\\ \tt{:}\dashrightarrow \boxed{\bf v=69.2m/s}$