355417
A particle is acted upon by a force \(F\) which varies with position \(x\) as shown in figure. If the particle at \(x=0\) has kinetic energy of 25 \(J\), then the kinetic energy of the particle at \(x = 16\,m\) is
1 45\(J\)
2 30\(J\)
3 70\(J\)
4 135\(J\)
Explanation:
Work done \(\mathrm{W}=\) Area under \(\mathrm{F}\) - \(\mathrm{x}\) graph. Area is \((+) v e\) if \(F\) is \((+) v e\). Area is \((-) v e\) if \(F\) is \((-)\) ve \(W=\left[\dfrac{1}{2} \times 6 \times 10\right]+4 \times[-5]+[4 \times 5]+[2 \times(-5)]\)\(\Rightarrow W=30-20+20-10=20 \mathrm{~J}\) According to work energy theorem, \(\begin{aligned}& K_{f}-K_{i}=W \\& K_{f}=25 J+20 J=45 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355418
If the linear momentum of a body is increased by \(50 \%\), then the kinetic energy of that body increases by
1 \(100 \%\)
2 \(125 \%\)
3 \(225 \%\)
4 \(25 \%\)
Explanation:
Kinetic energy of the body, \(\mathrm{K}=\dfrac{p^{2}}{2 m}\) where \(p\) is the linear momentum and \(m\) is the mass of the body. Since mass remains constant \(\therefore \mathrm{K} \propto \mathrm{p}^{2}\) As \({p^\prime } = p + \frac{{50}}{{100}}p = \frac{3}{2}p\) \(\therefore \dfrac{K^{\prime}}{K}=\left(\dfrac{p^{\prime}}{p}\right)^{2}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}\) \(\%\) Increase in the kinetic energy \(=\dfrac{K^{\prime}-K}{K} \times 100 \%\) \(=\left(\dfrac{K^{\prime}}{K}-1\right) \times 100 \%=\left(\dfrac{9}{4}-1\right) \times 100 \%=125 \%\)
KCET - 2010
PHXI06:WORK ENERGY AND POWER
355419
A stationary particle breaks into two parts of masses \(m_{A}\) and \(m_{B}\) which move with velocities \(v_{A}\) and \(v_{B}\) respectively. The ratio of their kinetic energies \(\left(K_{B}: K_{A}\right)\) is
1 \(m_{B} v_{B}: m_{A} v_{A}\)
2 \(v_{B}: v_{A}\)
3 \(1: 1\)
4 \(m_{B}: m_{A}\)
Explanation:
According the principle of conservation of momentum, they fly off in opposite directions with same momentum, as external force on the system is zero. \(\therefore m_{A} v_{A}=m_{B} v_{B}\) \(\Rightarrow \dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}}\) \(\quad \;(1)\) As \(K=\dfrac{p^{2}}{2 m}\) As \(p=\) same and hence \(K \propto \dfrac{1}{m}\) \(\dfrac{K_{B}}{K_{A}}=\dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}} \quad\) (using (1))
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355420
Assertion : If momentum of a body increases by \(50 \%\), its kinetic energy will increase by \(125 \%\). Reason : Kinetic energy is proportional to square root of velocity (for constant mass).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(p^{\prime}=p+0.5 p=1.5 p\) \(\Rightarrow E^{\prime}{ }_{k}=\left(\dfrac{1}{2} m v^{2}\right)=\dfrac{\left(p^{\prime}\right)^{2}}{2 m}=(2.25) \dfrac{p^{2}}{2 m}=2.25 E_{k}\) \(\Rightarrow \Delta E_{k}=1.25 E_{k}\) \(\Rightarrow\) Kinetic energy increases by \(125 \%\) Assertion is correct. As \(K . E \propto v^{2}\) Reason is wrong. So correct option is (3).
355417
A particle is acted upon by a force \(F\) which varies with position \(x\) as shown in figure. If the particle at \(x=0\) has kinetic energy of 25 \(J\), then the kinetic energy of the particle at \(x = 16\,m\) is
1 45\(J\)
2 30\(J\)
3 70\(J\)
4 135\(J\)
Explanation:
Work done \(\mathrm{W}=\) Area under \(\mathrm{F}\) - \(\mathrm{x}\) graph. Area is \((+) v e\) if \(F\) is \((+) v e\). Area is \((-) v e\) if \(F\) is \((-)\) ve \(W=\left[\dfrac{1}{2} \times 6 \times 10\right]+4 \times[-5]+[4 \times 5]+[2 \times(-5)]\)\(\Rightarrow W=30-20+20-10=20 \mathrm{~J}\) According to work energy theorem, \(\begin{aligned}& K_{f}-K_{i}=W \\& K_{f}=25 J+20 J=45 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355418
If the linear momentum of a body is increased by \(50 \%\), then the kinetic energy of that body increases by
1 \(100 \%\)
2 \(125 \%\)
3 \(225 \%\)
4 \(25 \%\)
Explanation:
Kinetic energy of the body, \(\mathrm{K}=\dfrac{p^{2}}{2 m}\) where \(p\) is the linear momentum and \(m\) is the mass of the body. Since mass remains constant \(\therefore \mathrm{K} \propto \mathrm{p}^{2}\) As \({p^\prime } = p + \frac{{50}}{{100}}p = \frac{3}{2}p\) \(\therefore \dfrac{K^{\prime}}{K}=\left(\dfrac{p^{\prime}}{p}\right)^{2}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}\) \(\%\) Increase in the kinetic energy \(=\dfrac{K^{\prime}-K}{K} \times 100 \%\) \(=\left(\dfrac{K^{\prime}}{K}-1\right) \times 100 \%=\left(\dfrac{9}{4}-1\right) \times 100 \%=125 \%\)
KCET - 2010
PHXI06:WORK ENERGY AND POWER
355419
A stationary particle breaks into two parts of masses \(m_{A}\) and \(m_{B}\) which move with velocities \(v_{A}\) and \(v_{B}\) respectively. The ratio of their kinetic energies \(\left(K_{B}: K_{A}\right)\) is
1 \(m_{B} v_{B}: m_{A} v_{A}\)
2 \(v_{B}: v_{A}\)
3 \(1: 1\)
4 \(m_{B}: m_{A}\)
Explanation:
According the principle of conservation of momentum, they fly off in opposite directions with same momentum, as external force on the system is zero. \(\therefore m_{A} v_{A}=m_{B} v_{B}\) \(\Rightarrow \dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}}\) \(\quad \;(1)\) As \(K=\dfrac{p^{2}}{2 m}\) As \(p=\) same and hence \(K \propto \dfrac{1}{m}\) \(\dfrac{K_{B}}{K_{A}}=\dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}} \quad\) (using (1))
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355420
Assertion : If momentum of a body increases by \(50 \%\), its kinetic energy will increase by \(125 \%\). Reason : Kinetic energy is proportional to square root of velocity (for constant mass).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(p^{\prime}=p+0.5 p=1.5 p\) \(\Rightarrow E^{\prime}{ }_{k}=\left(\dfrac{1}{2} m v^{2}\right)=\dfrac{\left(p^{\prime}\right)^{2}}{2 m}=(2.25) \dfrac{p^{2}}{2 m}=2.25 E_{k}\) \(\Rightarrow \Delta E_{k}=1.25 E_{k}\) \(\Rightarrow\) Kinetic energy increases by \(125 \%\) Assertion is correct. As \(K . E \propto v^{2}\) Reason is wrong. So correct option is (3).
355417
A particle is acted upon by a force \(F\) which varies with position \(x\) as shown in figure. If the particle at \(x=0\) has kinetic energy of 25 \(J\), then the kinetic energy of the particle at \(x = 16\,m\) is
1 45\(J\)
2 30\(J\)
3 70\(J\)
4 135\(J\)
Explanation:
Work done \(\mathrm{W}=\) Area under \(\mathrm{F}\) - \(\mathrm{x}\) graph. Area is \((+) v e\) if \(F\) is \((+) v e\). Area is \((-) v e\) if \(F\) is \((-)\) ve \(W=\left[\dfrac{1}{2} \times 6 \times 10\right]+4 \times[-5]+[4 \times 5]+[2 \times(-5)]\)\(\Rightarrow W=30-20+20-10=20 \mathrm{~J}\) According to work energy theorem, \(\begin{aligned}& K_{f}-K_{i}=W \\& K_{f}=25 J+20 J=45 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355418
If the linear momentum of a body is increased by \(50 \%\), then the kinetic energy of that body increases by
1 \(100 \%\)
2 \(125 \%\)
3 \(225 \%\)
4 \(25 \%\)
Explanation:
Kinetic energy of the body, \(\mathrm{K}=\dfrac{p^{2}}{2 m}\) where \(p\) is the linear momentum and \(m\) is the mass of the body. Since mass remains constant \(\therefore \mathrm{K} \propto \mathrm{p}^{2}\) As \({p^\prime } = p + \frac{{50}}{{100}}p = \frac{3}{2}p\) \(\therefore \dfrac{K^{\prime}}{K}=\left(\dfrac{p^{\prime}}{p}\right)^{2}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}\) \(\%\) Increase in the kinetic energy \(=\dfrac{K^{\prime}-K}{K} \times 100 \%\) \(=\left(\dfrac{K^{\prime}}{K}-1\right) \times 100 \%=\left(\dfrac{9}{4}-1\right) \times 100 \%=125 \%\)
KCET - 2010
PHXI06:WORK ENERGY AND POWER
355419
A stationary particle breaks into two parts of masses \(m_{A}\) and \(m_{B}\) which move with velocities \(v_{A}\) and \(v_{B}\) respectively. The ratio of their kinetic energies \(\left(K_{B}: K_{A}\right)\) is
1 \(m_{B} v_{B}: m_{A} v_{A}\)
2 \(v_{B}: v_{A}\)
3 \(1: 1\)
4 \(m_{B}: m_{A}\)
Explanation:
According the principle of conservation of momentum, they fly off in opposite directions with same momentum, as external force on the system is zero. \(\therefore m_{A} v_{A}=m_{B} v_{B}\) \(\Rightarrow \dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}}\) \(\quad \;(1)\) As \(K=\dfrac{p^{2}}{2 m}\) As \(p=\) same and hence \(K \propto \dfrac{1}{m}\) \(\dfrac{K_{B}}{K_{A}}=\dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}} \quad\) (using (1))
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355420
Assertion : If momentum of a body increases by \(50 \%\), its kinetic energy will increase by \(125 \%\). Reason : Kinetic energy is proportional to square root of velocity (for constant mass).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(p^{\prime}=p+0.5 p=1.5 p\) \(\Rightarrow E^{\prime}{ }_{k}=\left(\dfrac{1}{2} m v^{2}\right)=\dfrac{\left(p^{\prime}\right)^{2}}{2 m}=(2.25) \dfrac{p^{2}}{2 m}=2.25 E_{k}\) \(\Rightarrow \Delta E_{k}=1.25 E_{k}\) \(\Rightarrow\) Kinetic energy increases by \(125 \%\) Assertion is correct. As \(K . E \propto v^{2}\) Reason is wrong. So correct option is (3).
355417
A particle is acted upon by a force \(F\) which varies with position \(x\) as shown in figure. If the particle at \(x=0\) has kinetic energy of 25 \(J\), then the kinetic energy of the particle at \(x = 16\,m\) is
1 45\(J\)
2 30\(J\)
3 70\(J\)
4 135\(J\)
Explanation:
Work done \(\mathrm{W}=\) Area under \(\mathrm{F}\) - \(\mathrm{x}\) graph. Area is \((+) v e\) if \(F\) is \((+) v e\). Area is \((-) v e\) if \(F\) is \((-)\) ve \(W=\left[\dfrac{1}{2} \times 6 \times 10\right]+4 \times[-5]+[4 \times 5]+[2 \times(-5)]\)\(\Rightarrow W=30-20+20-10=20 \mathrm{~J}\) According to work energy theorem, \(\begin{aligned}& K_{f}-K_{i}=W \\& K_{f}=25 J+20 J=45 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355418
If the linear momentum of a body is increased by \(50 \%\), then the kinetic energy of that body increases by
1 \(100 \%\)
2 \(125 \%\)
3 \(225 \%\)
4 \(25 \%\)
Explanation:
Kinetic energy of the body, \(\mathrm{K}=\dfrac{p^{2}}{2 m}\) where \(p\) is the linear momentum and \(m\) is the mass of the body. Since mass remains constant \(\therefore \mathrm{K} \propto \mathrm{p}^{2}\) As \({p^\prime } = p + \frac{{50}}{{100}}p = \frac{3}{2}p\) \(\therefore \dfrac{K^{\prime}}{K}=\left(\dfrac{p^{\prime}}{p}\right)^{2}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}\) \(\%\) Increase in the kinetic energy \(=\dfrac{K^{\prime}-K}{K} \times 100 \%\) \(=\left(\dfrac{K^{\prime}}{K}-1\right) \times 100 \%=\left(\dfrac{9}{4}-1\right) \times 100 \%=125 \%\)
KCET - 2010
PHXI06:WORK ENERGY AND POWER
355419
A stationary particle breaks into two parts of masses \(m_{A}\) and \(m_{B}\) which move with velocities \(v_{A}\) and \(v_{B}\) respectively. The ratio of their kinetic energies \(\left(K_{B}: K_{A}\right)\) is
1 \(m_{B} v_{B}: m_{A} v_{A}\)
2 \(v_{B}: v_{A}\)
3 \(1: 1\)
4 \(m_{B}: m_{A}\)
Explanation:
According the principle of conservation of momentum, they fly off in opposite directions with same momentum, as external force on the system is zero. \(\therefore m_{A} v_{A}=m_{B} v_{B}\) \(\Rightarrow \dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}}\) \(\quad \;(1)\) As \(K=\dfrac{p^{2}}{2 m}\) As \(p=\) same and hence \(K \propto \dfrac{1}{m}\) \(\dfrac{K_{B}}{K_{A}}=\dfrac{m_{A}}{m_{B}}=\dfrac{v_{B}}{v_{A}} \quad\) (using (1))
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355420
Assertion : If momentum of a body increases by \(50 \%\), its kinetic energy will increase by \(125 \%\). Reason : Kinetic energy is proportional to square root of velocity (for constant mass).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(p^{\prime}=p+0.5 p=1.5 p\) \(\Rightarrow E^{\prime}{ }_{k}=\left(\dfrac{1}{2} m v^{2}\right)=\dfrac{\left(p^{\prime}\right)^{2}}{2 m}=(2.25) \dfrac{p^{2}}{2 m}=2.25 E_{k}\) \(\Rightarrow \Delta E_{k}=1.25 E_{k}\) \(\Rightarrow\) Kinetic energy increases by \(125 \%\) Assertion is correct. As \(K . E \propto v^{2}\) Reason is wrong. So correct option is (3).
