A force is applied over a particle which displaces it from its origin to the .The work done on the particle in joules is
-7
+7
+10
+10
B.
+7
Work done in displacing the particle
A body of mass 2m moving with velocity v makes a head on elastic collision with another body of mass m which is initially at rest. Loss of kinetic energy of the colliding body (mass 2m) is
1/9 of its initial kinetic energy
1/6 of its initial kinetic energy
8/9 of its initial kinetic energy
1/2 of its initial kinetic energy
C.
8/9 of its initial kinetic energy
Initial K.E of ball of mass 2m = K1
Collision is elastic so both K.E and momentum are conserved. Let velocities of balls are v1 and v2 after collision
So, KE is conserved
And, momentum is conserved
⇒ (2m)v + m(0) = 2m (v1) + mv2
⇒ 2v = 2v1 + v2 ........ (ii)
Put this value in Eq. (i), we get
So, final K.E of ball of mass 2m,
Hence, loss of K.E. of 1st ball
A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is
40 m/s
20 m/s
10 m/s
10 m/s
A.
40 m/s
The potential energy of a 1 kg particle free move along the x-axis is given by
The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is
2
B.
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
7.2 J
3.6 J
120 J
120 J
B.
3.6 J
Mass per length
= M/L
= 4/2 = 2 kg/m
The mass of 0.6 m of chain = 0.6 x 2 = 1.2 kg
The centre of mass of hanging part = 0.6 +0 /2 = 0.3 m
Hence, work done in pulling the chain on the table
W =mgh
= 1.2 x 10 x 0.3
= 3.6 J