Question

A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?

Answer 1

A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then

(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s

(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.

c) The speed of the 0.950- kg puck after the collision is 2.82 m/s

d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.

Given:

Mass of ice puck, m₁ = 0.350 kg

Mass of another puck, m₂ = 0.950 kg

Velocity of ice puck, v₁ = 5.22 m/s

Velocity of another puck, v₂ = 0 m/s

$v^{'}_1= ?$

$v^{'}_2= ?$

$m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }$

$v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }$

$(0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s$

Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.

$v^{'}_2 = v^{'} + v_1$

    $= -2.40+5.22$

    $= 2.82 m/s$

Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.

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