Question

Problem 7.26 A 0.250-kg ice puck, moving east with a speed of 5.62 m/s , has a head-on collision with a 0.900-kg puck initially at rest. Assume that the collision is perfectly elastic.

Part A What is the speed of the 0.250-kg puck after the collision?

Express your answer to three significant figures and include the appropriate units.

v1 =

Part B What is the direction of the velocity of the 0.250-kg puck after the collision? What is the direction of the velocity of the 0.250- puck after the collision?

a) to the east

b) to the west

Part C What ise the speed of the 0.900-kg puck after the collision?

Express your answer to three significant figures and include the appropriate units.

v2 =

Part D What is the direction of the velocity of the 0.900-kg puck after the collision? What is the direction of the velocity of the 0.900- puck after the collision?

a) to the east

b) to the west

Answer 1

a)

m₁ = 0.250 kg

v_1i = 5.62 m/s

v_1f = ?

m2 = 0.900 kg

v_2i = 0 m/s

v_2f = ?

using conservation of momentum

m₁ v_1i + m₂ v_2i = m₁ v_1f + m₂ v_2f

(0.250) (5.62) + (0.900) (0) = (0.25) v_1f + (0.9) V_2f

V_2f = (1.41 - (0.25) v_1f ) /0.9                                 eq-1

usingh conservation of kinetic energy

m₁ v²_1i + m₂ v²_2i = m₁ v²_1f + m₂ v²_2f

using eq-1

(0.250) (5.62)² + (0.900) (0)² = (0.25) v²_1f + (0.9) ((1.41 - (0.25) v_1f ) /0.9)²

v_1f = - 3.62 m/s

direction = west     since v_1f is negative so opposite to east

using eq-1

V_2f = (1.41 - (0.25) v_1f ) /0.9 = (1.41 - (0.25) (-3.62)) /0.9 = 2.6 m/s   towards east   since v_2f is positive