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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

A projectile of mass 1 . 2kg undergoes a perfectly inelastic...

A projectile of mass 1.2kg1 . 2kg undergoes a perfectly inelastic collision with a trolley of mass 3.6kg3 . 6kg as shown in the figure. At the time of the collision, the speed of the projectile is 5ms−15m{s}^{- 1} at an angle of 37°37 ° with the horizontal. The trolley is free to move, only along a horizontal rail which coincides with the direction of the projectile's horizontal motion. Assuming that the trolley doesn't topple, calculate the amount of heat energy (in JJ) released in the collision. [Take sin⁡37°=35\sin 37 ° = \dfrac{3}{5} and cos⁡37°=45\cos 37 ° = \dfrac{4}{5}]
A projectile of mass 1 . 2kg undergoes a perfectly inelastic collision with a trolley of mass 3 . 6kg as shown in the

Correct answer

12.6

Step-by-step solution

A projectile of mass 1 . 2kg undergoes a perfectly inelastic collision with a trolley of mass 3 . 6kg as shown in the

A projectile of mass 1 . 2kg undergoes a perfectly inelastic collision with a trolley of mass 3 . 6kg as shown in the

Using the conservation of linear momentum in the horizontal direction we get,

1.2×4=4.8×v1 . 2 \times 4 = 4 . 8 \times v

⇒v=1ms−1⇒ v = 1m{s}^{- 1}

The loss in kinetic energy is equal to the amount of heat energy released.

∣∆K∣=12×1.2×(5)2−12×4.8×(1)2\left|∆ K\right| = \dfrac{1}{2} \times 1 . 2 \times {\left(5\right)}^{2} - \dfrac{1}{2} \times 4 . 8 \times {\left(1\right)}^{2}

\left|∆ K\right| = 12 . 6J

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