Two masses of 1,kg and 16,kg are moving with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by the formula $P = \sqrt{2mE}$.Since the kinetic energy $E$

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(B) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by the formula $P = \sqrt{2mE}$.Since the kinetic energy $E$ is equal for both masses,we have $P \propto \sqrt{m}$.Therefore,the ratio of the linear momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.Given $m_1 = 1\,kg$ and $m_2 = 16\,kg$,we get $\frac{P_1}{P_2} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Thus,the ratio is $1:4$.

Two masses of $1\,kg$ and $16\,kg$ are moving with equal kinetic energy. The ratio of the magnitude of their linear momentum is:

Similar Questions

The graph between kinetic energy $E$ and velocity $v$ is:

If the linear momentum is increased by $5\%$,the kinetic energy will increase by.......$\%$

Two bodies with masses $M_1$ and $M_2$ have equal kinetic energies. If $p_1$ and $p_2$ are their respective momenta,then $p_1/p_2$ is equal to

If $E - V < 0$,is this condition possible?

The momentum of a body is increased by $50 \%$. The percentage increase in the kinetic energy of the body is $...........\,\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module