A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struc
ID: 1366853 • Letter: A
Question
A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 8.1 m/s. After the collision, the 0.20-kg puck has a speed of4.9 m/s at an angle of
= 53°
to the positive x-axis.
(a) Determine the velocity of the 0.30-kg puck after the collision.
(b) Find the fraction of kinetic energy lost in the collision.
= %
magnitude m/s directionYour response differs from the correct answer by more than 10%. Double check your calculations.° from the positive x-axis
Explanation / Answer
Answer:-
INITIAL SITUATION:
pi= m1u1+m2u2 = (0.20 kg)(8.1 m/s) + (0.30 kg)(0 m/s) = 1.62 kg m/s
Ki= (1/2)m1u1^2+(1/2)m2u2^2=(1/2)(0.20 kg)(8.1 m/s)2 +(1/2)(0.30 kg)(0 m/s)2 = 6.561 J
AFTER COLLISION:
Assuming the collision is elastic, then both momentum and kinetic energy is conserved.
Along x -axis
pi= pf
=> 1.62 kg m/s = (0.20 kg)(4.9 m/s)cos530 +(0.30 kg)v2cos(th)0
=> (0.30 kg)v2cos(th)0 = 1.62 kg m/s – 0.5897 kg m/s
=> v2cos(th)0 = 1.0302/0.30 = 3.434 --- (1)
Along y -axis
pi= pf
=> 0 = (0.20 kg)(4.9 m/s)sin530 - (0.30 kg)v2sin(th)0
=> (0.20 kg)(4.9 m/s)sin530 = (0.30 kg)v2sin(th)0
=> v2sin(th)0 = 0.7826 / 0.30 = 2.608 --- (2)
Add and sqrt 1 and 2 equations
v2^2 = 3.434^2+ 2.608^2
=>v2^2 = 17.008
=>v2= 4.124 m/s
2) E_initial=E_final + E_lost
E_lost=E_initial -E_final
Ei= 6.561 J
Ef= (1/2)m1v1^2+(1/2)m2v2^2
6.561 J = (1/2)m1v1^2+(1/2)m2v2^2 + E_lost
6.561 J = (1/2)(0.20 kg)(4.9 m/s)2 +(1/2)(0.30 kg)(4.124 m/s)2 + E_lost
E_lost = 6.561 J – 2.401 J – 2.551 J
E_lost = 1.609 J
E_lost/E_initial = 1.609 J / 6.561 J = 0.245 (24.5%)
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