Two bodies having same mass 40 kg are moving in opposite directions

• Correct Answer: D

  Solution :

           By the conservation of momentum \[40\times 10+(40)\times (-7)=80\times v\] Þ \[v=1.5\,m/s\]

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Page 2

• Correct Answer: D

  Solution :

  
  The momentum of third part will be equal and opposite to the resultant of momentum of rest two equal parts let V is the velocity of third part. By the conservation of linear momentum \[3m\times V=m\times 12\sqrt{2}\]Þ \[V=4\sqrt{2}\ m/s\]

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Page 3

• Correct Answer: A

  Solution :

  
  Particle falls from height h then formula for height covered by it in nth rebound is given by \[{{h}_{n}}=h{{e}^{2n}}\] where e = coefficient of restitution, n = No. of rebound Total distance travelled by particle before rebounding has stopped \[H=h+2{{h}_{1}}+2{{h}_{2}}+2{{h}_{3}}+2{{h}_{n}}+........\] \[=h+2h{{e}^{2}}+2h{{e}^{4}}+2h{{e}^{6}}+2h{{e}^{8}}+.........\] \[=h+2h({{e}^{2}}+{{e}^{4}}+{{e}^{6}}+{{e}^{8}}+.......)\] \[=h+2h\left[ \frac{{{e}^{2}}}{1-{{e}^{2}}} \right]=h\,\left[ 1+\frac{2{{e}^{2}}}{1-{{e}^{2}}} \right]=h\,\left( \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right)\]

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Page 4

• Correct Answer: D

  Solution :

           Due to the same mass of A and B as well as due to elastic collision velocities of spheres get interchanged after the collision.

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Page 5

• Correct Answer: A

  Solution :

  
  From the formulae \[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\ {{u}_{1}}\] We get \[v=\left( \frac{M-m}{M+m} \right)\ u\]

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Page 6

• Correct Answer: A

  Solution :

           Momentum conservation          \[5\times 10+20\times 0=5\times 0+20\times v\]Þ\[v=2.5\,m/s\]

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Page 7

• Correct Answer: D

  Solution :

           Due to elastic collision of bodies having equal mass, their velocities get interchanged.

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Page 8

• Correct Answer: C

  Solution :

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Page 9

• Correct Answer: B

  Solution :

           \[{{m}_{1}}=2\,kg\]and\[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}\]\[=\frac{{{u}_{1}}}{4}\](given)          By solving we get\[{{m}_{2}}=1.2\,kg\]

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Page 10

• Correct Answer: C

  Solution :

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Page 11

• Correct Answer: D

  Solution :

           It is clear from figure that the displacement vector \[\Delta \overrightarrow{r}\] between particles \[{{p}_{1}}\] and \[{{p}_{2}}\] is \[\Delta \overrightarrow{r}=\overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}}=-8\hat{i}-8\hat{j}\]
  
  \[|\Delta \overrightarrow{r}|\,=\sqrt{{{(-8)}^{2}}+{{(-8)}^{2}}}=8\sqrt{2}\]                        ?..(i) Now, as the particles are moving in same direction \[(\because \ \overrightarrow{{{v}_{1}}}\text{ and }\overrightarrow{{{v}_{2}}}\text{ are }+ve)\], the relative velocity is given by \[{{\overrightarrow{v}}_{rel}}=\overrightarrow{{{v}_{2}}}-\overrightarrow{{{v}_{1}}}=(\alpha -4)\hat{i}+4\hat{j}\] \[{{\overrightarrow{v}}_{rel}}=\sqrt{{{(\alpha -4)}^{2}}+16}\]                         ?..(ii) Now, we know \[|{{\overrightarrow{v}}_{rel}}|\,=\frac{|\Delta \overrightarrow{r}|}{t}\] Substituting the values of \[{{\overrightarrow{v}}_{rel}}\] and \[|\Delta \overrightarrow{r}|\] from equation (i) and (ii) and \[t=2s\], then on solving we get \[\alpha =8\]

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Page 12

• Correct Answer: B

  Solution :

                Fractional decrease in kinetic energy of neutron =\[1-{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\]  [As \[{{m}_{1}}=1\]and\[{{m}_{2}}=\text{ }2\]]          \[=1-{{\left( \frac{1-2}{1+2} \right)}^{2}}\]\[=1-{{\left( \frac{1}{3} \right)}^{2}}=1-\frac{1}{9}=\frac{8}{9}\]       

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Page 13

• Correct Answer: A

  Solution :

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Page 14

• Correct Answer: B

  Solution :

           When target is very light and at rest then after head on elastic collision it moves with double speed of projectile i.e. the velocity of body of mass m will be 2v.

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Page 15

• Correct Answer: A

  Solution :

           In head on elastic collision velocity get interchanged (if masses of particle are equal). i.e. the last ball will move with the velocity of first ball i.e 0.4 m/s

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Page 16

• Correct Answer: A

  Solution :

           By the principle of conservation of linear momentum,          \[Mv=m{{v}_{1}}+m{{v}_{2}}\]Þ\[Mv=0+(M-m){{v}_{2}}\]Þ\[{{v}_{2}}=\frac{Mv}{M-m}\]

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Page 17

• Correct Answer: A

  Solution :

           Since bodies exchange their velocities, hence their masses are equal so that \[\frac{{{m}_{A}}}{{{m}_{B}}}=1\]

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Page 18

• Correct Answer: D

  Solution :

           \[mgh=\] initial potential energy               \[mg{h}'=\] final potential energy after rebound               As 40% energy lost during impact \mgh'=60% of mgh          Þ \[h'=\frac{60}{100}\times h=\frac{60}{100}\times 10=6\,m\]

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Page 19

• Correct Answer: C

  Solution :

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Page 20

• Correct Answer: A

  Solution :

           Fractional loss \[=\frac{\Delta U}{U}=\frac{mg(h-{h}')}{mgh}\]\[=\frac{2-1.5}{2}=\frac{1}{4}\]

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Page 21

• Correct Answer: C

  Solution :

           \[\frac{\Delta K}{K}=\left[ 1-{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}} \right]=\left[ 1-{{\left( \frac{m-2m}{m+2m} \right)}^{2}} \right]\]\[=\frac{8}{9}\]          \[\Delta K\]=\[\frac{8}{9}K\]i.e. loss of kinetic energy of the colliding body is \[\frac{8}{9}\] of its initial kinetic energy.

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Page 22

• Correct Answer: D

  Solution :

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Page 23

• Correct Answer: A

  Solution :

                                          \[mgh=\frac{80}{100}\times mg\times 100\] Þ \[h=80\,m\]

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Page 24

• Correct Answer: A

  Solution :

           Let ball is projected vertically downward with velocity v from height h Total energy at point \[A=\frac{1}{2}m{{v}^{2}}+mgh\] During collision loss of energy is 50% and the ball rises up to same height. It means it possess only potential energy at same level. 50%\[\left( \frac{1}{2}m{{v}^{2}}+mgh \right)=mgh\] \[\frac{1}{2}\left( \frac{1}{2}m{{v}^{2}}+mgh \right)=mgh\] \[v=\sqrt{2gh}=\sqrt{2\times 10\times 20}\] \\[v=20\,m/s\]
  

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Page 25

• Correct Answer: A

  Solution :

                \[{{h}_{n}}=h{{e}^{2n}}\] after third collision \[{{h}_{3}}=h{{e}^{6}}\] [as n = 3]       

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Page 26

• Correct Answer: A

  Solution :

                                          Let mass A moves with velocity v and collides in elastically with mass B, which is at rest.
  
  According to problem mass A moves in a perpendicular direction and let the mass B moves at angle q with the horizontal with velocity v. Initial horizontal momentum of system (before collision)   = mv                                  ....(i) Final horizontal momentum of system (after collision)    \[~~=mVcos\theta \]                      ....(ii) From the conservation of horizontal linear momentum      mv = mV cosq Þ v = V cosq                   ...(iii) Initial vertical momentum of system (before collision) is zero. Final vertical momentum of system \[\frac{mv}{\sqrt{3}}-mV\sin \theta \] From the conservation of vertical linear momentum \[\frac{mv}{\sqrt{3}}-mV\sin \theta =0\]Þ\[\frac{v}{\sqrt{3}}=V\sin \theta \]                 ...(iv) By solving (iii) and (iv) \[{{v}^{2}}+\frac{{{v}^{2}}}{3}={{V}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\]   Þ \[\frac{4{{v}^{2}}}{3}={{V}^{2}}\]  Þ \[V=\frac{2}{\sqrt{3}}v\].

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