Question

Asteroid X, speeding at 100 m/s towards an equally massive but stationary asteroid Y, collides with a glancing blow. After collision, X is deflected at 35 degrees from its original direction while Y travels at 45 degrees to the original direction of X, assume total momentum is conserved.

a. Find the speed of each asteroid after collision.

b. What fraction of the original KE is lost in the collision?

Answer 1

Given initial velocities of masses m1 and m2 are equal m1=m2= m

let initial valeocities be U1 and U2 and final velocities be V1 and V2

   U1= 100m/s U2 = 0   

Momentum in all direction is conserved.

m1U1x +m2U2x = m1V1x +m2V2x

   m1U1y +m2U2y = m1V1y +m2V2y

m(100) +m(0) = m(V1 cos 35) +m(V2cos45)

   100 = 0.819 V1 + 0.707 V2

m1U1y +m2U2y = m1V1y +m2V2y

0 = V1y +V2y

   = V1 sin35 -V2 sin45

0.573 V1 = 0.707V2

V1 = 1.234 V2

  100 = 0.819 V1 + 0.707 V2

   100 = 0.819(1.234V2) + 0.707V2

100= 1.01 V2 + 0.707 V2

   100 = 1.717V2

   V2= 58.24 m/s

Speed of asteroid after collision V1= 1.234 V2 =1.234(5.24)=71.87 m/s

Kinetic energy lost in energy = 0.5 m(U1^2 - V1^2) + 0.5m( 0- v2^2) = 0.5M( 100^2 -71.87^2 -58.24^2)

ENERGY LOST = 721.4m J

Original kinetic energy =0.5m(100^2) = 5000m J

energy lost/ original kinetic energy = 721.4m /5000m =0.144