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Motion in a Straight Line

Questions and explanations focused on motion along a straight line. Covers displacement, velocity, acceleration, and interpreting motion graphs.

282 questions

A 0.6kg pendulum bob completes a vertical circle of radius 1.2m. What is the speed at the top? (Take g\=10m/s2)

At top, minimum speed vC=gL=10×1.2=12≈3.46m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 3 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

A 10kg mass at 8m/s collides inelastically with a stationary 5kg mass. What is the final speed?

Momentum conservation: 10×8=(10+5)vf⇒80=15vf⇒vf=8015≈5.33m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 5 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

A 9kg mass at 7m/s collides inelastically with a stationary 3kg mass. What is the final speed?

Momentum conservation: 9×7=(9+3)vf⇒63=12vf⇒vf=5.25m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 5 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

A 11kg mass at 6m/s collides inelastically with a stationary 4kg mass. What is the final speed?

Momentum conservation: 11×6=(11+4)vf⇒66=15vf⇒vf=4.4m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 4 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

A 1.8kg pendulum bob completes a vertical circle of radius 2m. What is the speed at the top? (Take g\=10m/s2)

At top, minimum speed vC=gL=10×2=20≈4.47m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 4 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

A 1.2kg pendulum bob completes a vertical circle of radius 1.6m. What is the speed at the top? (Take g\=10m/s2)

At top, minimum speed vC=gL=10×1.6=16=4m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 4 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0