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You and your friends are doing physics experiments on a frozen pond that serves

ID: 1485294 • Letter: Y

Question

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 79.0 kg , is given a push and slides eastward. Abigail, with mass 60.0 kg , is sent sliding northward. They collide, and after the collision Sam is moving at 36.0 north of east with a speed of 7.00 m/s and Abigail is moving at 15.0 south of east with a speed of 10.0 m/s . a)What was the speed of each person before the collision? Sam's speed: b)Abigail's speed: c)By how much did the total kinetic energy of the two people decrease during the collision?

Explanation / Answer

m1 =79 kg , u1y =0

m2 = 60 kg , u2x =0

v1x = 7cos(36) = 5.663

v1y = 7sin(36) = 4.1145

v2x = 10cos(15) = 9.66

v2y =-10sin(15) = - 2.588

From conservation of momentum along x - direction

m1u1x+m2u2x =m1v1x+m1v1y

(79*u1) +0 = (79* 5.663) + (60*9.66) ... (1)

From conservation of momentum along y direction

m1u1y+m2u2y =m1v1y +m2v2y

0 +(60*u2) = (79*4.1154 )-(60*2.588) ...(2)

By solving (1) and (2) we get

a) initila velocity of Sam u1 = 13 m/s

b) initial velcotiy of Abigali u2 = 2.83 m/s

c) Ki = (1/2)m1u1^2 +(1/2)m2u2^2

Ki = (0.5*79*13*13)+(0.5*60*2.83*2.83)

Ki =6915.8 J

Kf = (1/2)m1v1^2 +(1/2)m2v2^2

Kf = (0.5*79*7*7)+(0.5*60*10*10)

Kf =4935.5 J

Lost in total kinetic energy Ki -Kf = (6915.8 ) -(4935.5)

Lost in total kinetic energy =1980.3 J