Question

Same force acts on two bodies of masses $m_1$ and $m_2$ ($m_2$> $m_1$). After travelling the same distance,

• A. Kinetic energy of $m_2$ is greater than kinetic energy of $m_1$
• B. Kinetic energy of $m_1$ is greater than kinetic energy of $m_2$
• C. Ratio of kinetic energies of those two is equal to ratio of their momentum
• D. Kinetic energy of $m_1$ is equal to kinetic energy of $m_2$

Answer 1

Answer: (D)

$s=\frac{1}{2}a{t}^2$
Now, $K.E_1=\frac{1}{2}{m}_1(at{)}^2$$=\frac{1}{2}\frac{m_1}{a_1}s{a}_1=F.s$
$K.E_2=\frac{1}{2}mg(a_{2}t{)}^2$$=F.s$
i.e $K.E_1=K{E}_2$