SATHEE: Work Energy And Power Question 289

Work Energy And Power Question 289

Question: A block of mass 0.50 kg is moving with a speed of $2.00 m s^{- 1}$ on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Options:

A) 0.16 J

B) 1.00 J

C) 0.67 J

D) 0.34 J

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Initial kinetic energy of the system

$K . E_{i} = \frac{1}{2} m u^{2} + \frac{1}{2} M {(0)}^{2} = \frac{1}{2} \times 0.5 \times 2 \times 2 + 0 = 1 J$

For collision, applying conservation of linear momentum $m \times u = (m + M) \times v$

$∴ 0.5 \times 2 = (0.5 + 1) \times v \Rightarrow v = \frac{2}{3} m / s$

Final kinetic energy of the system is $K . E_{f} = \frac{1}{2} (m + M) v^{2} = \frac{1}{2} (0.5 + 1) \times \frac{2}{3} \times \frac{2}{3} = \frac{1}{3} J$

$∴$ Energy loss during collision $= (1 - \frac{1}{3}) J = 0.67 J$

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Work Energy And Power Question 289

Question: A block of mass 0.50 kg is moving with a speed of 2.00ms−1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Options:

Answer:

Solution:

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Question: A block of mass 0.50 kg is moving with a speed of $2.00 m s^{- 1}$ on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is