Question

A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.790 m/s . It has a head-on collision with a 0.310kg glider that is moving to the left with a speed of 2.10 m/s . Suppose the collision is elastic.

Part A Find the magnitude of the final velocity of the 0.149 kg glider. m/s SubmitMy AnswersGive Up Part B Find the direction of the final velocity of the 0.149 kg glider. Find the direction of the final velocity of the 0.149  glider. to the right to the left SubmitMy AnswersGive Up Part C Find the magnitude of the final velocity of the 0.310 kg glider. m/s SubmitMy AnswersGive Up Part D Find the direction of the final velocity of the 0.310 kg glider. Find the direction of the final velocity of the 0.310  glider. to the right to the left

Answer 1

m1 = 0.149 kg   m2 = 0.31 kg

speeds before collision

u1 = 0.79 m/s     u2 = -2.1 m/s

speeds after collision

v1 = ?        v2 = ?

initial momentum before collision

Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from moentum conservation

total momentum is conserved

Pf = Pi

m1*u1 + m2*u2 = m1*v1 + m2*v2

(0.149*0.79) - (0.31*2.1) = (0.149*v1) + (0.31*v2).....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2

KEi = KEf

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

0.5*0.149*0.79^2 + 0.5*0.31*2.1^2 = 0.5*0.149*v1^2 + 0.5*0.31*v2^2 ........(2)

v2 = 20.9 m/s <---------------answer

solving 1 & 2

v1 = -3.114 m/s

v2 = -0.224 m/s

part(A)

v1 = 3.114 m/s

part(B)

left

part(C)

v2 = 0.224 m/s

part(D)

left