AIIMS2019Physics-Work, Energy and Power

AIIMS 2019 Physics Collisions MCQ Question

Type: MCQ-numerical-Medium-Class 11

A body of mass 5×10³ kg moving with speed 2 m/s collides with a body of mass 15×10³ kg in elastically and sticks to it. Then loss in K.E. of the system will be:

A

7.5 kJ

B

15 kJ

C

10 kJ

D

5 kJ

Correct Answer

Option A

Detailed Explanation

In an inelastic collision, kinetic energy (K.E.) is not conserved, but momentum is. For the given bodies, the initial momentum is pi=(5×103kg×2m/s)+(15×103kg×0)=1×104kg m/sp_i = (5 \times 10^3 \, \text{kg} \times 2 \, \text{m/s}) + (15 \times 10^3 \, \text{kg} \times 0) = 1 \times 10^4 \, \text{kg m/s}. After the collision, the combined mass is 20×103kg20 \times 10^3 \, \text{kg}, and using momentum conservation, the final speed vf=pitotal mass=1×10420×103=0.5m/sv_f = \frac{p_i}{\text{total mass}} = \frac{1 \times 10^4}{20 \times 10^3} = 0.5 \, \text{m/s}. The initial K.E. is 12(5×103)(22)=10kJ\frac{1}{2}(5 \times 10^3)(2^2) = 10 \, \text{kJ} and the final K.E. is 12(20×103)(0.52)=2.5kJ\frac{1}{2}(20 \times 10^3)(0.5^2) = 2.5 \, \text{kJ}. Thus, the loss in K.E. is $ 10 , \text{kJ} - 2.5 , \text{kJ} =

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