VA ENG Phys chap g Mom x C Secure https: www.webassign Student Assignment-Respon
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VA ENG Phys chap g Mom x C Secure https: www.webassign Student Assignment-Responses submit?dep 16045795 A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle of 51.0 to the positive (see the figure below Before the collision After the collision vif sin vy cos vyf cos sin (a) Determine the velocity of the 0.300-kg puck after the collision. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. Need Help? Read It 342 PM Type here to search 6M 5/2017Explanation / Answer
(a) According to the conservation of momentum
Intial momentum = final momentum
Since after the collision the pucks have velocity component in both direction i.e x and y therefore we will conserve the momentum in x and y direction respectively
In x direction
Initial momentum = m1V1 + m2V2 = 0.2*2 + 0.3*0 = 0.4 kg-m/s
In y direction
Initial momentum = 0 (because the velocity of both masses were not in the direction of y)
Now final momentum
In x direction = m1V1X + m2V2X = 0.2*(1Cos51) + 0.3V2X
In y direction = m1V1Y + m2V2Y = 0.2*(1Sin51) + 0.3*V2Y
Now equating the initial and final momentum in x and y direction respectively
First in x direction
0.2*(1Cos51) + 0.3V2X = 0.4
V2X = 0.9138 m/s
Now in y direction
0.2*(1Sin51) + 0.3*V2Y = 0
V2Y = -0.518 m/s
Hence velocity of the puck 2 = (V2X 2 + V2Y2)1/2 = 1.05 m/s
(b) Now the energy lost = Initial energy - final energy
Intial energy = (1/2)m1V12 + (1/2)m2V22 = 0.5*(0.2)*22 + 0.5*0.3*0 = 0.4 J
Final energy = 1/2)m1V1F2 + (1/2)m2V2F2 = 0.5*0.2*12 + 0.5*0.3*1.052 = 0.2654 J
Energy lost = 0.4 - 0.2654 = 0.134 J
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