Question

Two bodies, one of mass 4 kg and the other of mass 6 kg, approach each other head-on with speeds of 2 m/s and 8 m/s, respectively. Calculate the velocity of each body after collision if the collision is perfectly elastic.​

Answer 1

Answer:

J=20 N

Step-by-step explanation:

ANSWER

Applying conservation of linear momentum in x direction

u1=2 and u2=3 (in −x direction)

initial momentum =m2−m3

final momentum =mv2−mv1

equating both

−m=mv2−mv1

⟹v2−v1=−1 ..............eq(1)

from definition of coeficient of restitution which states that

velocity of separation=e×(velocity of approach)

⟹v2−(−v1)=u1−(−u2)

v2+v1=2+3=5...............eq(2)

solving eq(1) and eq(2), we get

v1=3 and v2=2 

considering ball B

J(impulse)=mv2−m(−u2)=m(v2+u2)=4(2+3)

J=20N