Question

12 6. In a nuclear reaction carbon nucleus, C collides elastically in a head on collision with an unknown nucleus. The initial speed of the carbon nucleus is 110 km/h and the speed of the unknown nucleus is 400km/h. After the collision the carbon nucleus moves in opposite direction to its original direction with a speed of 145km/h. i) Determine the ratio of the mass of Carbon nucleus and the unknown nucleus. ii) What is the velocity of the unknown nucleus after the collision?

Answer 1

mass of carbon = m1

mass of unknown = m2

before collision

speed of m1 , v1i = 110 km/h

speed of m2 , v2i = -400 km/h

after collision

speed of m1 , v1if = -145 km/h

speed of m2 , v2f = 0 km/h

from momentum conservation

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i - (2*m2*v2i))/(m1+m2)

-145 = ((m1 - m2)*110 - (2*m2*400))/(m1 + m2)

-145*m1 - 145*m2 = 110*m1 - 110*m2 - 800*m2

m1*(110 + 145 ) = m2*(800 - 145 + 110)

m1/m2 = 3 <<<-----ANSWER

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

v2f = (-(m2-3m2)*400 +(2*3m2*110))/(3m2 + m2)

v2f = (800 + 660)/4

v2f = 365 km/h <<----------ANSWER