Answer:
(a) V (A) = 0.7 m/s,
(b) V (A) = 0.7 m/s,
(c) V (B) = 0.7 m/s
(d) u= - 0.60 m/s
(e) v = 0.75 m/s
Explanation:
Given:
M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s
Sol:
a) law of conservation of momentum
M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)
so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V
after solving V = 0.7 m/s
After the collision the velocities of the both block will be as the the spring is compressed maximum.
V (A) = 0.7 m/s
b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )
c) as stated above the in the Part (a)
V(B) = 0.7 m/s
d) When the both blocks moved apart after the collision:
Let u=velocity of block A after the collision.
and v = velocity of block B after the collision.
then conservation of momentum
M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u
⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v
⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)
For elastic collision
the velocity relative approach = velocity relative separation
so 2.00 m/s = v-u ----- eqn (2)
⇒v = u + 2.00 m/s
putting this value in eqn (1) we get
2.00 m/s = u + 1.86 (v + 2.00 m/s)
u= - 0.60 m/s
e) putting v= 2.00 m/s in eqn (1)
2.00 m/s = - 2.32 m/s + 1.86 v
v = 0.75 m/s
