If the position of a moving particle, with respect to time be, x(t) = 1 + t − t2, then, the acceleration of particle is given by (assume all measurements in MKS)
− 1 ms-2
− 1.5 ms-2
− 2 ms-2
− 2.5 ms-2
A ball is dropped from a tower of height h. The duration (t0) of motion, when it reaches bottom of the tower is given by
For the given position-time (x − t) graph , the interval in which velocity is zero, is
(t3 − t1)
(t3 − t2)
(t5 − t4)
(t5 − t2)
A particle moves along the x-axis as x = u (t −2s) + a (t − 2s)2. The initial velocity of the particle is
u − 2a
u − 4a
2a − u
2a − 3u
The centre of mass of three particles of mass m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.0 kg at the corners of an equilateral triangle 1.0 m on a side, as shown in figure
A system consists of two point masses M and m (< M). The centre of mass of the system is
at the middle of m and M
nearer to M
nearer to m
at the position of large mass
A ball is projected horizontally with a velocity of 5 m/s from the top of a building 19.6 m high, flow long will the ball take to hit the ground?
2 s
3 s
A body of mass 3 kg is acted on by a force which varies as shown in the graph. The momentum acquired is given by
zero
5 N-s
30 N-s
50 N-s
Two perfectly elastic particles A and B of equal masses travelling along the line joining, them with velocity 15 m/s and 20 m/s respectively collide. Their velocities after the elastic collision will be (in m/s) respectively.
0 and 25
5 and 20
10 and 15
20 and 15
A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms-1 . The kinetic energy of the other mass is
192 J
96 J
144 J
288 J
D.
288 J
By law of conservation of linear momentum, the velocity of the 4 kg mass is 12 ms-1. Therefore,
KE =