20 Physics -- Work, Energy and Power

The blocks A and B have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.1 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) the collision is perfectly elastic and (b) the...

The blocks A and B have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.1 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. g=10 m/s2 .

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a) In a lab frame of reference,

Velocity of center of mass is given by,

v'=(mv_OA+ mv_OB)/2m

Now, in center of mass frame of reference,

Since, there will be perfectly elastic collision. After collision, they exchange their velocity.

In lab frame of reference,

Deceleration due to the friction = µg

From kinematics we have,

V_F=V_B- µgt

Since the final velocity of B is zero, we get t=10s.

Distance travelled d is given by,

d = V_Bt - (µgt²)/2

Inserting given values, the distance travelled by block is 50m.

b) After perfectly inelastic collision, both blocks A and B travel as a combined mass.

m(V_A+V_B)=2mV_AB

We get, initial velocity of the combined mass 5m/s.

The friction of the block B will have affect whole body.

Then we have,

V_F=V_AB- µgt

d = V_ABt - (µgt²)/2

Solving the equations, the distance travelled by the body is 12.5m.

Work-Energy Theorem can also be used to find the distance.

A man cycles up a hill, whose slope is 1 in 20 with a velocity of 6.4km/hr along the hill. The weight of the man and cycle is 98kg. What work per minute is he doing? What is his horsepower?

A cord is used to lower vertically a block of mass M, through a distance "d" at a constant downward acceleration of g/8. The work done by the cord on the block isa. Mgd b. Mgd/8c. 3Mgd/8d. -7Mgd/8

Here, Ma=Mg-T

or, M(g/8)=Mg-T

or, T= M(g-g/8)

or, T= 7Mg/8

so, work done = T*d cos180⁰

=7Mg/8 * (-d)

= -7Mgd/8

So, correct option is d

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The change in magnitude of the ball is 28 kgm/s.

Solution:

Here,

mass of the ball (m) = 2kg

initial velocity of the ball (u) = 8 m/s (say downward direction is positive)

final velocity of the ball (v) = -6 m/s (because in upward direction)

Hence,

change in linear momentum of the ball (delta P) = m(v-u)

= 2 (-6 - 8)

= 2 (-14)

= -28 kgm/s

The magnitude of change in linear momentum of the ball is 28 kgm/s.

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