Collision in 2 Dimensions

Question

Remember that when dealing with collisions in 2-dimensions, momentum is conserved in each dimension.

Two pucks are sliding on a frictionless tabletop. Block A (mass 4.76 kg) is moving to the East (+x direction) at 1.90 m/s. Block B (mass 5.71 kg) is moving to the North (+y direction) at 2.28 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.)

a)What is the x-component of the total momentum of the system before the collision?

b)What is the y-component of the total momentum of the system before the collision?

c)What is the x-component of the total momentum of the system after the collision?

d)What is the y-component of the total momentum of the system after the collision?

Answer 1

Momentum = M.v

The direction of Momentum is the same as that of the velocity. Therefore, the questions above are reduced to finding the x and y-components of velocity of each puck before, and after collision.

BEFORE COLLISION:

Block A

velocity x-component U_x^A = 1.90 m/s

velocity y-component U_y^A= 0

Block B

velocity x-component U_x^B = 0

velocity y-component U_y^B = 2.28 m/s

AFTER COLLISION:

Block A

velocity x-component = V_x^A

velocity y-component = V_y^A

Block B

velocity x-component = V_x^B

velocity y-component = V_y^B

Use principle of conservation of total Momentum of the system (block A + block B) in x and y directions:

M_A*V_x^A + M_B*V_x^B = 1.9

M_A*V_y^A + M_B*V_y^B = 2.28

Note that if you wanted to determine final velocity components for both masses A and B, you will need two additional equations because there are 4 unknowns. The above question does not ask you to calculate these velocity components. So, the above answer is complete.

You can obtain additional equations using conservation of energy principle and then use them to solve for all 4 unknowns.