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Q.

A $2000 kg$ truck moving with $10 m / s$ strikes a car waiting at a traffic light. After the collision, the two move together at $8 m / s$ . The mass of the car is

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a

250 kg

b

500 kg

c

750 kg

d

1000 kg

answer is B.

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Detailed Solution

The mass of the car is

500 kg

.
The type of collision which there is a loss of kinetic energy and after the collision both the colliding particles move together is called a perfectly inelastic collision. In this type of collision, the coefficient of restitution is equal to zero. The momentum of the system remains constant in this collison.
Question Image

Question Image

Momentum before collision (

P_{1}

)

=

Momentum after the collision (

P_{2}

)

P_{1} = m_{1} u_{1} + m_{2} u_{2}

P_{2} = m_{1} v_{1} + m_{2} v_{2}

Here, speed of the car is

0 m / s

After the collision speed is

8 m / s

According to the principle of conservation of momemtum

8 = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}

Where

m_{1}

is the mass of truck,

v_{1}

is the velocity of truck,

v_{2}

is the velocity of car and

m_{2}

is the mass of car.

\Rightarrow

8 = \frac{2000 \times 10 + m_{2} \times 0}{2000 + m_{2}}

\Rightarrow 16000 + 8 m_{2} = 20000

\Rightarrow

m_{2} = \frac{4000}{8} kg

\Rightarrow

m_{2} = 500 kg

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