Question

Problem 7.26 A 0.250-kg ice puck, moving east with a speed of 5.62 m/s , has...

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

a)

m1 = 0.250 kg

v1i = 5.62 m/s

v1f = ?

m2 = 0.900 kg

v2i = 0 m/s

v2f = ?

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(0.250) (5.62) + (0.900) (0) = (0.25) v1f + (0.9) V2f

V2f = (1.41 - (0.25) v1f ) /0.9                                 eq-1

usingh conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

using eq-1

(0.250) (5.62)2 + (0.900) (0)2 = (0.25) v21f + (0.9) ((1.41 - (0.25) v1f ) /0.9)2

v1f = - 3.62 m/s

direction = west     since v1f is negative so opposite to east

using eq-1

V2f = (1.41 - (0.25) v1f ) /0.9 = (1.41 - (0.25) (-3.62)) /0.9 = 2.6 m/s   towards east   since v2f is positive

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