Q: A ball falling freely from a height of 4.9 ~m / s , hits a horizontal surface. If e= 3 4 , then the ball will hit the surface, second time after

aligned Velocity on hitting the surface & = 2 9.8 4.9 \\ & =9.8 ~m / s aligned Velocity after first bounce, v= 3 4 9.8 Time taken from first bounce to the second bounce = 2 v g =2 3 4 9.8 1 9.8 =1.5 ~s

Q: Two bodies are having kinetic energies in the ratio 16: 9. If they have same linear momentum, the ratio of their masses respectively is:

aligned & K.E. = p ^2 2 ~m \\ & K _1 ~K _2 = p _1^2 2 ~m _1 2 ~m _2 p _2^2 = m _2 ~m _1 = 16 9 \\ & m _1 ~m _2 = 9 16 aligned

Question 1

A ball falling freely from a height of 4.9 ~m / s , hits a horizontal surface. If e= 3 4 , then the ball will hit the surface, second time after

Accepted Answer

aligned Velocity on hitting the surface & = 2 9.8 4.9 \ & =9.8 ~m / s aligned Velocity after first bounce, v= 3 4 9.8 Time taken from first bounce to the second bounce = 2 v g =2 3 4 9.8 1 9.8 =1.5 ~s

Question 2

Two bodies of mass 1 kg and 3 kg have position vectors i +2 j +k and -3 i -2 j + k respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of v

Accepted Answer

Position of COM of a mass - system is given as, r _ com = m_1 r _1+m_2 r _2 m_1+m_2 = 1( i +2 j + k )+3(-3 i -2 j + k 1+3 =-2 i - j + k | r _ c m |=|-2 i - j + k |= (2)^2+(1)^2+(1)^2 = 6 Only option (1) magnitude is 1^2+(-2)^2+1^2 = 6 So option (1) is correct.

Question 3

Two bodies are having kinetic energies in the ratio 16: 9. If they have same linear momentum, the ratio of their masses respectively is:

Accepted Answer

aligned & K.E. = p ^2 2 ~m \ & K _1 ~K _2 = p _1^2 2 ~m _1 2 ~m _2 p _2^2 = m _2 ~m _1 = 16 9 \ & m _1 ~m _2 = 9 16 aligned

Question 4

A particle of mass m moving with velocity v collides with a stationary particle of mass 2 ~m . After collision, they stick together and continue to move together with velocity

Accepted Answer

Applying conservation of linear momentum P _ i = P _ f ( P = mv ) m v_1+2 m v_2=(m+2 m) v^ m v^1+2 m 0=(3 m) v^ mv =3 mv ^ v ^ = v 3

Question 5

Three bricks each of length L and mass M are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is

Accepted Answer

x_ 1 = L 2 , x_ 2 = L , x_ 3 = 5 ~L 4 x_ CM = m_ 1 x_ 1 +m_ 2 x_ 2 +m_ 3 x_ 3 m_ 1 +m_ 2 +m_ 3 = M L 2 + M L + M 5 ~L 4 M + M + M = 11 ~L 12

Center of Mass Momentum and Collision — Previous Year Questions & Solutions