Questions and solutions focused on collisions between objects and calculations involving power in mechanical systems.
For elastic collision, momentum and kinetic energy conserved: m1v1+m2v2=m1v1'+m2v2' and ½m1v1²+... per NCERT. Solving gives velocities matching 0 m/s. Inelastic loses energy. So option A is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
At top, v=gL, energy at bottom = energy at top. As per NCERT, applying relevant law/formula with correct units and sign convention leads to √75 m/s = 5√3 m/s ≈ 8.66 m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Potential energy mgh=5×10×9=450J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 13 m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Kinetic energy K=½mv² and potential U=mgh per NCERT. Energy conservation K+U=constant for conservative forces. Substituting m, v, h gives 778 m/s as energy, matching option B.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives -7.55 J. Work-energy theorem W=ΔK confirms. So option B is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Power P=W/t=F·v per NCERT. Rate of doing work. With given force and velocity, P=Fv. Calculation yields 84 kW. Hence option B correctly represents power.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Power P=W/t=F·v per NCERT. Rate of doing work. With given force and velocity, P=Fv. Calculation yields 160 W. Hence option B correctly represents power.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives -1.27 J. Work-energy theorem W=ΔK confirms. So option B is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Work W=60×5=300J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 300 J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Kinetic energy K=½mv² and potential U=mgh per NCERT. Energy conservation K+U=constant for conservative forces. Substituting m, v, h gives 6 J as energy, matching option B.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Initial Ki=12×10×12=5J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 420 J. Work-energy theorem W=ΔK confirms. So option B is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Collisions and Power
