JEE Physics System Of Particles and Rotational Motion MCQs Set 55

Answer: (a) \( \left(\dfrac7{12},\dfrac{\sqrt3}4\right) \) m

Explanation:
1. The third vertex is at \( \left(0.5,\ \dfrac{\sqrt3}2\right) \).
2. \( x=\dfrac{50(0)+100(1)+150(0.5)}{300}=\dfrac7{12} \) m.
3. \( y=\dfrac{150\cdot\sqrt3/2}{300}=\dfrac{\sqrt3}4 \) m.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) \( \left(\dfrac{5a}{12},\dfrac{5b}{12}\right) \)

Explanation:
1. The sheet has its CM at \( (a/2,b/2) \). The removed quarter has mass \( M/4 \) with CM at \( (3a/4,3b/4) \).
2. \( x=\dfrac{M(a/2)-\tfrac M4(3a/4)}{3M/4}=\dfrac{5a}{12} \), and similarly \( y=\dfrac{5b}{12} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) \( \dfrac a5(\hat i-\hat j) \)

Explanation:
1. \( \vec a_{cm}=\dfrac{\sum m\vec a}{\sum m} \) with \( \sum m=10m \).
2. x: \( \dfrac{-m+3m}{10m}a=\dfrac a5 \). y: \( \dfrac{2m-4m}{10m}a=-\dfrac a5 \).
3. \( \vec a_{cm}=\dfrac a5(\hat i-\hat j) \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) \( \tan\theta=\dfrac13 \)

Explanation:
1. Take B at the origin, A at \( (0,l) \), C at \( (l,0) \). The CM of two equal rods is at \( (l/4,l/4) \).
2. At equilibrium the line from A to the CM is vertical. This line makes an angle \( \theta \) with AB where \( \tan\theta=\dfrac{l/4}{l-l/4}=\dfrac13 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( \dfrac{3h}4 \)

Explanation:
1. The CM of a solid cone lies on the axis at a distance \( h/4 \) from the base.
2. From the vertex, \( z_0=h-\dfrac h4=\dfrac{3h}4 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) \( \dfrac a{10} \)

Explanation:
1. Total mass \( =6m+m+m+m+m=10m \).
2. \( y_{cm}=\dfrac{6m(0)+m(a)+m(a)+m(0)+m(-a)}{10m}=\dfrac a{10} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) 10 m/s

Explanation:
1. \( v_{cm}=\dfrac{10(14)+4(0)}{14}=10 \) m/s, and it stays constant since no external horizontal force acts.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) zero

Explanation:
1. The only forces are internal, so the momentum of the system stays zero.
2. Hence the centre of mass remains at rest.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) A is false, but R is true.

Explanation:
1. Statement I is false: the velocity of the CM is governed by the net external force, not the torque; zero torque only means the angular momentum about the CM is constant.
2. Statement II is true: with no external force the total linear momentum is conserved.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) (a) and (b) only

Explanation:
1. Momentum: \( mv=MV \). Energy: \( mgR=\tfrac12mv^2+\tfrac12MV^2 \), giving \( v=\sqrt{\dfrac{2gR}{1+m/M}} \): (a) true and (d) false.
2. The horizontal CM does not move: \( m\,\Delta x_m+M\,\Delta x_M=0 \) with \( \Delta x_m-\Delta x_M=R \), so \( \Delta x_M=-\dfrac{mR}{M+m} \): (b) true.
3. The mass is then at \( x=-R+\dfrac{MR}{M+m}=-\dfrac{mR}{M+m} \), with no \( \sqrt2 \): (c) false.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) 0.5 m/s

Explanation:
1. No external force acts, so the velocity of the centre of mass is constant.
2. Hence it remains 0.5 m/s (it does not scale with the relative velocity).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) \( \left(L+2R,0\right) \)

Explanation:
1. The small sphere's centre moves \( 2(6R-R)=10R \) relative to the large sphere, towards the left.
2. The system's CM stays fixed: \( 4M\,\Delta x+M(\Delta x-10R)=0\Rightarrow\Delta x=2R \).
3. So the large sphere's centre moves to \( (L+2R,0) \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) 6.53 s

Explanation:
1. Relative speed 49 m/s, gap 98 m: they meet at \( t=2 \) s at height \( 98-19.6=78.4 \) m. Velocities: first \( 49-19.6=29.4 \) up, second \( 19.6 \) down.
2. Combined velocity \( =\dfrac{29.4-19.6}2=4.9 \) m/s upward. Then \( -78.4=4.9t-4.9t^2 \) gives \( t^2-t-16=0 \), \( t=\dfrac{1+\sqrt{65}}2=4.53 \) s.
3. Total \( =2+4.53=6.53 \) s.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) 9 cm, on the side away from the hole

Explanation:
1. R = 28 cm, r = 21 cm; the hole's centre is \( 28-21=7 \) cm from the plate's centre. Mass ratio \( =r^2/R^2=9/16 \).
2. \( x=-\dfrac{(9/16)(7)}{1-9/16}=-9 \) cm, i.e. 9 cm from the centre, away from the hole.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) \( Mgl\,(1+\theta_0^2) \)

Explanation:
1. Work done = gain in PE + gain in KE. PE gain \( =Mgl \).
2. Angular momentum is conserved: \( v'=v\dfrac L{L-l} \), so \( \Delta KE=\tfrac12Mv^2\left[\dfrac{L^2}{(L-l)^2}-1\right]\approx\dfrac{Mv^2l}{L} \).
3. With \( v^2=gL\theta_0^2 \) at the bottom, \( \Delta KE=Mgl\theta_0^2 \). Total \( =Mgl(1+\theta_0^2) \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) \( \dfrac{2v^2}{5g} \)

Explanation:
1. At maximum height the particle and wedge move with a common velocity: \( mv=(m+4m)V\Rightarrow V=v/5 \).
2. \( \tfrac12mv^2=\tfrac12(5m)\dfrac{v^2}{25}+mgh\Rightarrow h=\dfrac{v^2}{2g}\cdot\dfrac45=\dfrac{2v^2}{5g} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( 2\sqrt{\dfrac pk} \)

Explanation:
1. Impulse = change in momentum: \( \int_0^Tkt\,dt=\dfrac{kT^2}2=3p-p=2p \).
2. \( T=2\sqrt{p/k} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.