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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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If a particle of mass

m_1

approaches another particle of mass

m_2

, initially at rest, with speed

v_v

, the speed of the

\mathrm{cm}

of the system is

v_{\mathrm{cm}}=\frac{m_1 v_1}{m_1+m_2}

before collision takes place. If the collision is elastic, the speed of the

\mathrm{cm}

of the system after the collision is

View answer and explanation

Correct option: B

\frac{m_1 v_1}{m_1+m_2}

Option B (

\frac{{m_1 v_1}}{{m_1+m_2}}

) is correct because in an elastic collision with no external forces, the velocity of the center of mass remains constant.

More Questions from WORK-ENERGY-POWER

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A ball throwing cannon, lifts ball and throws with

72 \mathrm{~km} / \mathrm{hr}

Speed. If balls are raised upto

10 \mathrm{~m}

. then find Input power of cannon if it lifts 8 balls per minute and its efficiency is

80 \%

. Mass of each ball is

1 \mathrm{~kg}

. Assertion:When a particle moves in a circle with a uniform speed its acceleration is constant but the velocity changes.
Reason: Angular displacement is not an axial vector. A body of mass

m

thrown vertically upwards attains a maximum height

h

. As what height will its kinetic energy be

75 \%

of its initial value? Assertion: Centripetal force does no work.

Reason: Force and displacement are perpendicular to each other. Statement I: Both stretched and compressed springs possess potential energy.

Statement II: Work done against restoring force is stored as potential energy. A bullet of mass

m

moving at a speed

v

hits a ball of mass

M

kept at rest. A small part having mass

m^{\prime}

breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed

v_1

in the direction of the bullet. Find the velocity of the bullet after the collision.