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A ball is projected from ground with a velocity V at an angle θ to the vertical. On its path it makes an elastic collision with a vertical wall & returns to ground. The total time of flight of the ball is:
A particle is projected from a smooth horizontal surface with velocity V at an angle θ from horizontal. Coefficient of restitution between surface & ball is e. The distance of the point where ball strikes the surface second time from the point of projection is ?
An isolated particle of mass m is moving in horizontal plane (x-y) along the axis at certain height above the ground. It suddenly explodes into two fragments of masses 3/d & 3m/4 . An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at:
m1y1 + m2y2 – O
= y2 = –5 cm
A ball of mass 0.2 kg rests on a vertical post of height 5m. A bullet of mass 0.01 kg, travelling with a velocity v m/s in horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m & the bullet at a distance of 100 m from the foot of the post. The initial velocity v of the bullet is:
0.01 V = 0.24 + 0.01 * 5u
Time of flight (T) = 1s
Range of ball = u × T = 4 × 1 = u = 20 m/s
V = 500 m/s
Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are V & 2V respectively as shown in the figure. Between collisions, the particle move with constant speeds. After making how many elastic collisions, other than that A, these two particles will again reach the point A?
An impulse J1 is imparted when a cricket ball is hitted with a bat by a person. The person hits the ball again but with half the force. Also the bat & ball are in contact for half of the time as in the first case. Then new impulse is J2. Then:
A bullet of mass 100 g is fired from below into the bob of mass 400 g of a long simple pendulum as shown in figure. The bullet remains inside the bob & the bob rises through a height of 5m. The speed of bullet is (Take g = 10 m/s2):
By principle of conservation of linear momentum
(0.1)V + 0 = (0.1 + 0.4)V'
= V' = V/5
Now by using equation V2 = u2 + 2gh
= we get V = 50 m/s
be the velocity of the system then internal forces can change:
Internal forces can not change velocity but can do work.
Ball 1 and 2 of masses 4 kg & 2 kg respectively are moving towards each other with velocities 10 m/s & 5 m/s & undergo a head on collision. After the collision, the velocity of ball 2 is observed to be 7 m/s to the right then coefficient of restitution is:
A thick uniform wire is bent into the shape of a letter "C". Then its centre of mass lies at a position nR from the straight line joining the two end of the wire. The value of 'n' is:
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