A car of mass $1200 k g$ moving at a speed of $72 k m h^{- 1}$ collides with another car of mass $1500 k g$ which is at rest. What is the common velocity of the two
cars, if they move together after the collision (m s–1)?

$32.04$

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$3.204$

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$8.9$

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$89$

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Solution

The correct option is C
$8.9$

The explanation for the correct answer
Step 1: Given
$m_{1} = 1200 kg, m_{2} = 1500 kg, v_{1} = 72 k m h^{- 1} = 20 m s^{- 1}, v_{2} = 0$
Let v be the common velocity after collision.
Step 2: Calculation
By conservation of linear momentum
Momentum before = Momentum after-------------1
Momentum before $= m_{1} v_{1} + m_{2} v_{2} = 1200 \times 20 + 1500 \times 0 = 24000 {kgms}^{- 1}$ --2)
Momentum after = $(m_{1} + m_{2}) v = (1200 + 1500) v = 2700 v {kgms}^{- 1}$ --3)
Substituting 2,3 in equation 1)
$24000 = 2700 v v = 8.9 m s^{- 1}$

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A car of mass 1200kg moving at a speed of 72kmh-1collides with another car of mass 1500kgwhich is at rest. What is the common velocity of the twocars, if they move together after the collision (m s–1)?

A car of mass $1200 k g$ moving at a speed of $72 k m h^{- 1}$ collides with another car of mass $1500 k g$ which is at rest. What is the common velocity of the two
cars, if they move together after the collision (m s–1)?