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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1

Two bodies of masses

m

and

2 m

are moving along positive

x

and

y

axes respectively with equal speeds of

4 {~m} / {s}

. They collide at origin and stick together. The final velocity of combined mass is

View answer and explanation

Correct option: A

\left(\frac{4}{3} \hat{i}+\frac{8}{3} \hat{j}\right) {m} / {s}

Option A (

\left(\frac{{4}}{{3}} \hat{{i}}+\frac{{8}}{{3}} \hat{{j}}\right) {m} / {s}

) is correct because it is derived from the conservation of momentum in both x and y directions during the collision.

More Questions from WORK-ENERGY-POWER

Practice All Chapter MCQs →

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.