Question

Two pucks are sliding on a frictionless tabletop. Block A (mass 4.11 kg) is moving to the East (+x direction) at 1.50 m/s. Block B (mass 4.93 kg) is moving to the North (+y direction) at 1.80 m/s. Assume the system to be both Block A and Block B. (Note: It could be useful to make a drawing of the situation.)
What is the x-component of the total momentum of the system before the collision?
What is the y-component of the total momentum of the system before the collision?
What is the x-component of the total momentum of the system after the collision?
What is the y-component of the total momentum of the system after the collision?

Answer 1

M_a = mass of block A = 4.11 kg           , M_b = mass of block B = 4.93 kg

Velocity of M_a along X-direction= V_axi = 1.50 m/s   ,

Velocity of M_b along X-direction before collision = V_bxi = 0 m/s

Velocity of M_a along Y-direction= V_ayi = 0 m/s   ,

Velocity of M_b along Y-direction before collision = V_byi = 1.80 m/s

Along X-direction :-

Total Momentum before Collision Along X-direction :

M_a V_axi + M_b V_bxi = (4.11 kg ) (1.50 m/s) + (4.93 kg) (0 m/s) = 6.165 kgm/s

Total Momentum before Collision Along Y-direction :

M_a V_axi + M_b V_bxi = (4.11 kg ) (0) + (4.93 kg) (1.80 m/s) = 8.87 kgm/s

According to Conservation of Momentum :

Total Momentum after Collision Along X-direction = Total Momentum before Collision Along X-direction = 6.165 kg m/s

Total Momentum after Collision Along Y-direction = Total Momentum before Collision Along Y-direction = 8.87 kgm/s