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lisabon 2012 [21]
3 years ago
13

The left end of a rod of length and rotational inertia is attached to a frictionless horizontal surface by a frictionless pivot,

as shown above. Point C marks the center (midpoint) of the rod. The rod is initially motionless but is free to rotate around the pivot. A student will slide a disk of mass toward the rod with velocity 0 perpendicular to the rod, and the disk will stick to the rod a distance x from the pivot. The student wants the rod-disk system to end up with as much angular speed as possible

Physics
1 answer:
natima [27]3 years ago
6 0

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

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The power in an electrical current is given by the equation
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2 years ago
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
Vedmedyk [2.9K]

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

5 0
3 years ago
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