If the momentum of a body of mass m is increa | vedclass

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

Question

(A) The kinetic energy $(K.E.)$ of a body is related to its momentum $(p)$ by the formula: $K.E. = \frac{p^2}{2m}$.When the momentum increases by $20

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) The kinetic energy $(K.E.)$ of a body is related to its momentum $(p)$ by the formula: $K.E. = \frac{p^2}{2m}$.When the momentum increases by $20 \%$,the new momentum $p'$ becomes:$p' = p + 0.20p = 1.2p$.The new kinetic energy $(K.E.')$ is:$K.E.' = \frac{(p')^2}{2m} = \frac{(1.2p)^2}{2m} = \frac{1.44p^2}{2m}$.Since $K.E. = \frac{p^2}{2m}$,we have $K.E.' = 1.44 	imes K.E$.The percentage increase in kinetic energy is given by:$\Delta K.E. \% = \frac{K.E.' - K.E.}{K.E.} 	imes 100 \%$$= \frac{1.44 K.E. - K.E.}{K.E.} 	imes 100 \%$$= 0.44 	imes 100 \% = 44 \%$.

If the momentum of a body of mass $m$ is increased by $20 \%$,then its kinetic energy increases by: (in $\%$)

Similar Questions

Two bodies of masses $m_1$ and $m_2$ are moving with the same kinetic energy. If $P_1$ and $P_2$ are their respective momenta,the ratio $\frac{P_1}{P_2}$ is equal to:

If $p$ is the linear momentum of an object and $v$ is its velocity,write the formula for its kinetic energy.

If the momentum of a body increases by $0.01\%$,its kinetic energy will increase by ........... $\%$

$A$ particle of mass $m_1$ moves with velocity $v_1$ and another particle of mass $m_2$ moves with velocity $v_2$. Both have the same momentum,but their kinetic energies are $E_1$ and $E_2$ respectively. If $m_1 > m_2$,then:

The kinetic energy of a body of mass $2 \ kg$ and momentum of $2 \ Ns$ is ............. $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module