Question

3. A 39 kg mass traveling to the right at a speed of 0.91c collides with a 47 kg mass also traveling to the right at a speed of 0.73c and sticks to it. How fast will the two masses be traveling after the collision? How much (kinetic) energy is lost in the collision?

Answer 1

3)

let m1 = 39 kg,

v1 = 0.91 c

m2 = 47 kg

v2 = 0.73*c

let V is the final speed of both blocks.

Apply conservation of momentum

(m1 + m2)*V = m1*v1 + m2*v2

V = (m1*v1 + m2*v2)/(m1+m2)

= (39*0.91*c + 47*0.73*c)/(39 + 47)

= 0.8116*c <<<<<<<------Answer

kinetic energy lost in the collision,

= (1/2)*m1*v1^2 + (1/2)*m2*v2^2 - (1/2)*(m1+m2)*V^2

= (1/2)*39*(0.91*3*10^8)^2 + (1/2)*47*(0.73*3*10^8)^2 - (1/2)*(39 + 47)*(0.8116*3*10^8)^2

= 3.13*10^16 J <<<<<<<------Answer