145809
The linear momentum \(p\) of a body varies with times as \(p=\alpha+\beta t^{2}\) where \(\alpha\) and \(\beta\) are constants. The net force action on the body for one dimensional motion varies as
1 \(t^{2}\)
2 \(\mathrm{t}^{-1}\)
3 \(\mathrm{t}^{-2}\)
4 \(\mathrm{t}\)
Explanation:
D Given, Momentum \((p)=\alpha+\beta t^{2}\) We know force in terms of momentum \(\text { Force }(F)=\frac{d p}{d t}\) \(F =\frac{d\left(\alpha+\beta t^{2}\right)}{d t}\) \(F =2 \beta t\) \(F \propto t\)
SCRA-2012
Laws of Motion
145810
The \(X\) and \(Y\) components of a force \(F\) acting at \(30^{\circ}\) to \(\mathrm{x}\)-axis are respectively:
D The \(\mathrm{X}\) component of force \((\mathrm{F})\) is \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \cos 30^{\circ}\) \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \mathrm{~F}\) The \(\mathrm{Y}\) component of force ' \(\mathrm{F}\) ' is \(\mathrm{F}_{\mathrm{Y}}=\mathrm{F} \sin 30^{\circ}=\mathrm{F} \times \frac{1}{2}=\frac{\mathrm{F}}{2}\)
Karnataka CET-2012
Laws of Motion
145811
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \(12 \mathrm{~m} / \mathrm{s}\). If the mass of the ball is \(0.15 \mathrm{~kg}\) the impulse imparted to the ball is
1 \(36 \mathrm{~N} \mathrm{~s}\)
2 \(3.6 \mathrm{~N} \mathrm{~s}\)
3 \(0.36 \mathrm{~N} \mathrm{~s}\)
4 \(0.036 \mathrm{~N} \mathrm{~s}\)
Explanation:
B Given, Initial velocity \(\left(\mathrm{v}_{\mathrm{i}}\right)=12 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(\mathrm{v}_{\mathrm{f}}\right)=-12 \mathrm{~m} / \mathrm{s}\) and mass \((\mathrm{m})=0.15 \mathrm{~kg}\) Initial momentum \(\left(\mathrm{p}_{\mathrm{i}}\right)=\mathrm{mv}_{\mathrm{i}}=0.15 \times 12=1.8 \mathrm{kgm} / \mathrm{sec}\) Final momentum \(\begin{aligned}\left(\mathrm{p}_{\mathrm{f}}\right) =\mathrm{mv}_{\mathrm{f}} \\ =0.15 \times(-12)=-1.8 \mathrm{kgm} / \mathrm{s}\end{aligned}\) Change in momentum \(\Delta \mathrm{p}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{\mathrm{i}}\) \(=-1.8-1.8\) \(=-3.6 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(|\Delta \mathrm{p}|= 3.6 \mathrm{kgms}^{-1}\) Then, \(\text { Impulse (I) }=|\Delta \mathrm{p}|\) \(\mathrm{I}=3.6 \mathrm{Ns}\)
J and K CET- 2011
Laws of Motion
145813
A cricket ball of mass \(0.5 \mathrm{~kg}\) strikes a cricket bat normally with a velocity of \(20 \mathrm{~m} \mathrm{~s}^{-1}\) and rebounds with a velocity of \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The impulse of the force exerted by the ball on the bat is
1 \(15 \mathrm{~N} \mathrm{~s}\)
2 \(25 \mathrm{~N} \mathrm{~s}\)
3 \(30 \mathrm{~N} \mathrm{~s}\)
4 \(10 \mathrm{~N} \mathrm{~s}\)
Explanation:
A Given mass \(=0.5 \mathrm{~kg}\) Initial velocity \(\left(v_{i}\right)=20 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(v_{f}\right)=-10 \mathrm{~m} / \mathrm{s}\) Change in momentum \((\Delta \mathrm{p})=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(\therefore \quad \Delta \mathrm{p}=0.5 \times(-10)-0.5 \times 20\) \(\Delta \mathrm{p}=-0.5(10+20)\) \(\Delta \mathrm{p}=-0.5 \times 30\) \(\Delta \mathrm{p}=-15 \mathrm{~N}\) \(\therefore \quad\) Impulse I \(=|\Delta \mathrm{p}|\) \(=15 \mathrm{Ns}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Laws of Motion
145809
The linear momentum \(p\) of a body varies with times as \(p=\alpha+\beta t^{2}\) where \(\alpha\) and \(\beta\) are constants. The net force action on the body for one dimensional motion varies as
1 \(t^{2}\)
2 \(\mathrm{t}^{-1}\)
3 \(\mathrm{t}^{-2}\)
4 \(\mathrm{t}\)
Explanation:
D Given, Momentum \((p)=\alpha+\beta t^{2}\) We know force in terms of momentum \(\text { Force }(F)=\frac{d p}{d t}\) \(F =\frac{d\left(\alpha+\beta t^{2}\right)}{d t}\) \(F =2 \beta t\) \(F \propto t\)
SCRA-2012
Laws of Motion
145810
The \(X\) and \(Y\) components of a force \(F\) acting at \(30^{\circ}\) to \(\mathrm{x}\)-axis are respectively:
D The \(\mathrm{X}\) component of force \((\mathrm{F})\) is \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \cos 30^{\circ}\) \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \mathrm{~F}\) The \(\mathrm{Y}\) component of force ' \(\mathrm{F}\) ' is \(\mathrm{F}_{\mathrm{Y}}=\mathrm{F} \sin 30^{\circ}=\mathrm{F} \times \frac{1}{2}=\frac{\mathrm{F}}{2}\)
Karnataka CET-2012
Laws of Motion
145811
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \(12 \mathrm{~m} / \mathrm{s}\). If the mass of the ball is \(0.15 \mathrm{~kg}\) the impulse imparted to the ball is
1 \(36 \mathrm{~N} \mathrm{~s}\)
2 \(3.6 \mathrm{~N} \mathrm{~s}\)
3 \(0.36 \mathrm{~N} \mathrm{~s}\)
4 \(0.036 \mathrm{~N} \mathrm{~s}\)
Explanation:
B Given, Initial velocity \(\left(\mathrm{v}_{\mathrm{i}}\right)=12 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(\mathrm{v}_{\mathrm{f}}\right)=-12 \mathrm{~m} / \mathrm{s}\) and mass \((\mathrm{m})=0.15 \mathrm{~kg}\) Initial momentum \(\left(\mathrm{p}_{\mathrm{i}}\right)=\mathrm{mv}_{\mathrm{i}}=0.15 \times 12=1.8 \mathrm{kgm} / \mathrm{sec}\) Final momentum \(\begin{aligned}\left(\mathrm{p}_{\mathrm{f}}\right) =\mathrm{mv}_{\mathrm{f}} \\ =0.15 \times(-12)=-1.8 \mathrm{kgm} / \mathrm{s}\end{aligned}\) Change in momentum \(\Delta \mathrm{p}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{\mathrm{i}}\) \(=-1.8-1.8\) \(=-3.6 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(|\Delta \mathrm{p}|= 3.6 \mathrm{kgms}^{-1}\) Then, \(\text { Impulse (I) }=|\Delta \mathrm{p}|\) \(\mathrm{I}=3.6 \mathrm{Ns}\)
J and K CET- 2011
Laws of Motion
145813
A cricket ball of mass \(0.5 \mathrm{~kg}\) strikes a cricket bat normally with a velocity of \(20 \mathrm{~m} \mathrm{~s}^{-1}\) and rebounds with a velocity of \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The impulse of the force exerted by the ball on the bat is
1 \(15 \mathrm{~N} \mathrm{~s}\)
2 \(25 \mathrm{~N} \mathrm{~s}\)
3 \(30 \mathrm{~N} \mathrm{~s}\)
4 \(10 \mathrm{~N} \mathrm{~s}\)
Explanation:
A Given mass \(=0.5 \mathrm{~kg}\) Initial velocity \(\left(v_{i}\right)=20 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(v_{f}\right)=-10 \mathrm{~m} / \mathrm{s}\) Change in momentum \((\Delta \mathrm{p})=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(\therefore \quad \Delta \mathrm{p}=0.5 \times(-10)-0.5 \times 20\) \(\Delta \mathrm{p}=-0.5(10+20)\) \(\Delta \mathrm{p}=-0.5 \times 30\) \(\Delta \mathrm{p}=-15 \mathrm{~N}\) \(\therefore \quad\) Impulse I \(=|\Delta \mathrm{p}|\) \(=15 \mathrm{Ns}\)
145809
The linear momentum \(p\) of a body varies with times as \(p=\alpha+\beta t^{2}\) where \(\alpha\) and \(\beta\) are constants. The net force action on the body for one dimensional motion varies as
1 \(t^{2}\)
2 \(\mathrm{t}^{-1}\)
3 \(\mathrm{t}^{-2}\)
4 \(\mathrm{t}\)
Explanation:
D Given, Momentum \((p)=\alpha+\beta t^{2}\) We know force in terms of momentum \(\text { Force }(F)=\frac{d p}{d t}\) \(F =\frac{d\left(\alpha+\beta t^{2}\right)}{d t}\) \(F =2 \beta t\) \(F \propto t\)
SCRA-2012
Laws of Motion
145810
The \(X\) and \(Y\) components of a force \(F\) acting at \(30^{\circ}\) to \(\mathrm{x}\)-axis are respectively:
D The \(\mathrm{X}\) component of force \((\mathrm{F})\) is \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \cos 30^{\circ}\) \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \mathrm{~F}\) The \(\mathrm{Y}\) component of force ' \(\mathrm{F}\) ' is \(\mathrm{F}_{\mathrm{Y}}=\mathrm{F} \sin 30^{\circ}=\mathrm{F} \times \frac{1}{2}=\frac{\mathrm{F}}{2}\)
Karnataka CET-2012
Laws of Motion
145811
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \(12 \mathrm{~m} / \mathrm{s}\). If the mass of the ball is \(0.15 \mathrm{~kg}\) the impulse imparted to the ball is
1 \(36 \mathrm{~N} \mathrm{~s}\)
2 \(3.6 \mathrm{~N} \mathrm{~s}\)
3 \(0.36 \mathrm{~N} \mathrm{~s}\)
4 \(0.036 \mathrm{~N} \mathrm{~s}\)
Explanation:
B Given, Initial velocity \(\left(\mathrm{v}_{\mathrm{i}}\right)=12 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(\mathrm{v}_{\mathrm{f}}\right)=-12 \mathrm{~m} / \mathrm{s}\) and mass \((\mathrm{m})=0.15 \mathrm{~kg}\) Initial momentum \(\left(\mathrm{p}_{\mathrm{i}}\right)=\mathrm{mv}_{\mathrm{i}}=0.15 \times 12=1.8 \mathrm{kgm} / \mathrm{sec}\) Final momentum \(\begin{aligned}\left(\mathrm{p}_{\mathrm{f}}\right) =\mathrm{mv}_{\mathrm{f}} \\ =0.15 \times(-12)=-1.8 \mathrm{kgm} / \mathrm{s}\end{aligned}\) Change in momentum \(\Delta \mathrm{p}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{\mathrm{i}}\) \(=-1.8-1.8\) \(=-3.6 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(|\Delta \mathrm{p}|= 3.6 \mathrm{kgms}^{-1}\) Then, \(\text { Impulse (I) }=|\Delta \mathrm{p}|\) \(\mathrm{I}=3.6 \mathrm{Ns}\)
J and K CET- 2011
Laws of Motion
145813
A cricket ball of mass \(0.5 \mathrm{~kg}\) strikes a cricket bat normally with a velocity of \(20 \mathrm{~m} \mathrm{~s}^{-1}\) and rebounds with a velocity of \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The impulse of the force exerted by the ball on the bat is
1 \(15 \mathrm{~N} \mathrm{~s}\)
2 \(25 \mathrm{~N} \mathrm{~s}\)
3 \(30 \mathrm{~N} \mathrm{~s}\)
4 \(10 \mathrm{~N} \mathrm{~s}\)
Explanation:
A Given mass \(=0.5 \mathrm{~kg}\) Initial velocity \(\left(v_{i}\right)=20 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(v_{f}\right)=-10 \mathrm{~m} / \mathrm{s}\) Change in momentum \((\Delta \mathrm{p})=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(\therefore \quad \Delta \mathrm{p}=0.5 \times(-10)-0.5 \times 20\) \(\Delta \mathrm{p}=-0.5(10+20)\) \(\Delta \mathrm{p}=-0.5 \times 30\) \(\Delta \mathrm{p}=-15 \mathrm{~N}\) \(\therefore \quad\) Impulse I \(=|\Delta \mathrm{p}|\) \(=15 \mathrm{Ns}\)
145809
The linear momentum \(p\) of a body varies with times as \(p=\alpha+\beta t^{2}\) where \(\alpha\) and \(\beta\) are constants. The net force action on the body for one dimensional motion varies as
1 \(t^{2}\)
2 \(\mathrm{t}^{-1}\)
3 \(\mathrm{t}^{-2}\)
4 \(\mathrm{t}\)
Explanation:
D Given, Momentum \((p)=\alpha+\beta t^{2}\) We know force in terms of momentum \(\text { Force }(F)=\frac{d p}{d t}\) \(F =\frac{d\left(\alpha+\beta t^{2}\right)}{d t}\) \(F =2 \beta t\) \(F \propto t\)
SCRA-2012
Laws of Motion
145810
The \(X\) and \(Y\) components of a force \(F\) acting at \(30^{\circ}\) to \(\mathrm{x}\)-axis are respectively:
D The \(\mathrm{X}\) component of force \((\mathrm{F})\) is \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \cos 30^{\circ}\) \(\mathrm{F}_{\mathrm{X}}=\mathrm{F} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \mathrm{~F}\) The \(\mathrm{Y}\) component of force ' \(\mathrm{F}\) ' is \(\mathrm{F}_{\mathrm{Y}}=\mathrm{F} \sin 30^{\circ}=\mathrm{F} \times \frac{1}{2}=\frac{\mathrm{F}}{2}\)
Karnataka CET-2012
Laws of Motion
145811
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \(12 \mathrm{~m} / \mathrm{s}\). If the mass of the ball is \(0.15 \mathrm{~kg}\) the impulse imparted to the ball is
1 \(36 \mathrm{~N} \mathrm{~s}\)
2 \(3.6 \mathrm{~N} \mathrm{~s}\)
3 \(0.36 \mathrm{~N} \mathrm{~s}\)
4 \(0.036 \mathrm{~N} \mathrm{~s}\)
Explanation:
B Given, Initial velocity \(\left(\mathrm{v}_{\mathrm{i}}\right)=12 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(\mathrm{v}_{\mathrm{f}}\right)=-12 \mathrm{~m} / \mathrm{s}\) and mass \((\mathrm{m})=0.15 \mathrm{~kg}\) Initial momentum \(\left(\mathrm{p}_{\mathrm{i}}\right)=\mathrm{mv}_{\mathrm{i}}=0.15 \times 12=1.8 \mathrm{kgm} / \mathrm{sec}\) Final momentum \(\begin{aligned}\left(\mathrm{p}_{\mathrm{f}}\right) =\mathrm{mv}_{\mathrm{f}} \\ =0.15 \times(-12)=-1.8 \mathrm{kgm} / \mathrm{s}\end{aligned}\) Change in momentum \(\Delta \mathrm{p}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{\mathrm{i}}\) \(=-1.8-1.8\) \(=-3.6 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) \(|\Delta \mathrm{p}|= 3.6 \mathrm{kgms}^{-1}\) Then, \(\text { Impulse (I) }=|\Delta \mathrm{p}|\) \(\mathrm{I}=3.6 \mathrm{Ns}\)
J and K CET- 2011
Laws of Motion
145813
A cricket ball of mass \(0.5 \mathrm{~kg}\) strikes a cricket bat normally with a velocity of \(20 \mathrm{~m} \mathrm{~s}^{-1}\) and rebounds with a velocity of \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The impulse of the force exerted by the ball on the bat is
1 \(15 \mathrm{~N} \mathrm{~s}\)
2 \(25 \mathrm{~N} \mathrm{~s}\)
3 \(30 \mathrm{~N} \mathrm{~s}\)
4 \(10 \mathrm{~N} \mathrm{~s}\)
Explanation:
A Given mass \(=0.5 \mathrm{~kg}\) Initial velocity \(\left(v_{i}\right)=20 \mathrm{~m} / \mathrm{s}\) Final velocity \(\left(v_{f}\right)=-10 \mathrm{~m} / \mathrm{s}\) Change in momentum \((\Delta \mathrm{p})=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(\therefore \quad \Delta \mathrm{p}=0.5 \times(-10)-0.5 \times 20\) \(\Delta \mathrm{p}=-0.5(10+20)\) \(\Delta \mathrm{p}=-0.5 \times 30\) \(\Delta \mathrm{p}=-15 \mathrm{~N}\) \(\therefore \quad\) Impulse I \(=|\Delta \mathrm{p}|\) \(=15 \mathrm{Ns}\)
