If a lighter body (mass $M_1$ and velocity $V_1$) and a heavier body (mass $M_2$ and velocity $V_2$) have the same kinetic energy,then

  • A
    $M_2 V_2 < M_1 V_1$
  • B
    $M_2 V_2 = M_1 V_1$
  • C
    $M_2 V_1 = M_1 V_2$
  • D
    $M_2 V_2 > M_1 V_1$

Explore More

Similar Questions

Answer the following:
$(a)$ The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
$(b)$ Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
$(c)$ An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance,however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
$(d)$ In Figure $(i)$ the man walks $2\; m$ carrying a mass of $15\; kg$ on his hands. In Figure $(ii)$,he walks the same distance pulling the rope behind him. The rope goes over a pulley,and a mass of $15\; kg$ hangs at its other end. In which case is the work done greater?

$A$ bullet of mass $10 \,g$ is fired horizontally with a velocity $1000 \,ms^{-1}$ from a rifle situated at a height $50 \,m$ above the ground. If the bullet reaches the ground with a velocity $500 \,ms^{-1}$, the work done against air resistance in the trajectory of the bullet is : $(g=10 \,ms^{-2})$ (in $\,J$)

$A$ wedge of mass $M = 4m$ lies on a frictionless plane. $A$ particle of mass $m$ approaches the wedge with speed $v$. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by

$A$ free body of mass $8 \ kg$ is moving at $2 \ m \cdot s^{-1}$ along a straight line. It splits into two equal parts due to an internal explosion releasing $16 \ J$ of energy,and neither part deviates from the original line of motion. What happens to the two parts?

Blocks of masses $m, m, 2m, 4m$ and $8m$ are arranged in a line on a frictionless floor. Another block of mass $m$,moving with speed $v$ along the same line (see figure) collides with the first mass $m$ in a perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass $8m$ starts moving,the total energy loss is $p\%$ of the original energy. The value of $p$ is close to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo