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)

m_{1} = 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

m_{1} v_{1i} + m_{2} v_{2i} =
m_{1} v_{1f} + m_{2} 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

m_{1} v^{2}_{1i} + m_{2}
v^{2}_{2i} = m_{1}
v^{2}_{1f} + m_{2}
v^{2}_{2f}

using eq-1

(0.250) (5.62)^{2} + (0.900) (0)^{2} = (0.25)
v^{2}_{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

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