SATHEE: Work Energy And Power Question 155

Work Energy And Power Question 155

Question: A mass of 10 gm moving with a velocity of 100 cm/s strikes a pendulum bob of mass 10 gm. The two masses stick together. The maximum height reached by the system now is $(g = 10 m / s^{2})$ [MP PET 1993]

Options:

A) Zero

B) 5 cm

C) 2.5 cm

D) 1.25 cm

Show Answer

Answer:

Correct Answer: D

Solution:

Initially mass 10 gm moves with velocity 100 cm/s \

Initial momentum = 10 × 100 = $1000 \frac{g m \times m}{\sec}$

After collision system moves with velocity $v_{s y s .}$

then Final momentum = $(10 + 10) \times v_{s y s .}$

By applying the conservation of momentum 10000 = $20 \times v_{s y s .}$

therefore $v_{s y s .} == 50 c m / s$

If system rises upto height h then $h = \frac{v_{s y s .}^{2}}{2 g} = \frac{50 \times 50}{2 \times 1000} = \frac{2.5}{2} = 1.25 c m$

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Work Energy And Power Question 155

Question: A mass of 10 gm moving with a velocity of 100 cm/s strikes a pendulum bob of mass 10 gm. The two masses stick together. The maximum height reached by the system now is (g=10m/s2) [MP PET 1993]

Options:

Answer:

Solution:

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Question: A mass of 10 gm moving with a velocity of 100 cm/s strikes a pendulum bob of mass 10 gm. The two masses stick together. The maximum height reached by the system now is $(g = 10 m / s^{2})$ [MP PET 1993]