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3 years ago

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Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of block A.a) Find the velocity of block A when the energy stored in the spring bumpers is maximum.b) Find the velocity of block A when the energy stored in the spring bumpers is maximum.c) Find the velocity of block B when the energy stored in the spring bumpers is maximum.d) Find the velocity of block A after they have moved apart.e) Find the velocity of block B after they have moved apart.

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

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

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