A particle starting from rest falls from a certain height.Assuming that the acceleration due to gravity remain the same throughout the motion, its displacements in three successive half second intervals are S₁, S₂ and S₃ then

S₁: S₂: S₃= 1: 5 : 9
S₁: S₂: S₃ = 1 : 3 : 5
S₁: S₂: S₃= 9: 2 : 3
S₁: S₂: S₃ = 1 : 1 : 1

Solution

Two bodies begin a free fall from the same height at a time interval of N s. If vertical separation between the two bodies is 1 after n second from the start of the first body, then n is equal to

\(\sqrt{nN}\)
\(\frac{1}{gN}\)
\(\frac{1}{gN}\)
\(\frac{1}{gN}-\frac{N}{2}\)

Solution

A ball is thrown vertically upward with a velocity ‘u’ from the balloon descending with velocity v.The ball will pass by the balloon after time

\(\frac{u-v}{2g}\)
\(\frac{u-v}{2g}\)
\(\frac{u-v}{2g}\)
\(\frac{u-v}{2g}\)

Solution

When two bodies move uniformly towards each other, the distance decreases by 6ms^-1. If both bodies move in the same directions with the same speeds (as above), the distance between them increases by 4 ms^-1. Then the speeds of the two bodies are

3ms^-1 and 3 ms^-1
4 ms^-1 and 2 ms^-1
5ms^-1 and 1 ms^-1
7 ms^-1 and 3 ms^-1

Solution

Let v_A and v_B are the velocities of two bodies.

In first case, v_A + v_B= 6m/s........(1)

In second case, v_A – v_B= 4m/s.........(2)

From (1) & (2) we get, v_A = 5 m/s and v_B=1 m/s.

A body starts from rest and travels a distance x with uniform acceleration, then it travels a distance 2x with uniform speed,finally it travels a distance 3x with uniform retardation and comes to rest. If the complete motion of the particle is along a straight line, then the ratio of its average velocity to maximum velocity is

Solution

A steel ball is bouncing up and down on a steel plate with a period of oscillation of 1 second. If g = 10 ms^-2, then it bounces up to a height of

5 m
10 m
2.5 m
1.25 m

Solution

Figure shows the position of a particle moving along the X-axis as a function of time

Which of the following is correct?

The particle has come to the rest 6 times
The maximum speed is at t = 6 s.
The velocity remains positive for t = 0 to t= 6 s.
The average velocity for the total period shown is negative.

Solution

At six points in the graph the tangents have zero slope i.e. velocity is zero.

In the displacement d versus time t graph given below,the value of average velocity in the time interval 0 to 20 s is(in m/s)

Solution

At t = 20s, d = 20 m

A graph of acceleration versus time of a particle starting from rest at t =0 is as shown in Fig. The speed of the particle at t = 14 second is

2ms^–1
34 ms^–1
20 ms^–1
42 ms^–1

Solution

Area under a-t graph is change in velocity.

Area=¹⁄₂ (4 × 4) + 6 × 4 + ¹⁄₂ × 2 × 4-¹⁄₂ × 2 × 2

=36 - 2 = 34 ms^–1

As initial velocity is zero therefore, the velocity at 14 second is 34 ms^–1

Two stones are thrown from the top of a tower, one straight down with an initial speed u and the second straight up with the same speed u. When the two stones hit the ground,they will have speeds in the ratio

2 : 3
2 : 1
1 : 2
1 : 1

Solution

Use v² – u² = 2aS. In both the cases, (u positive or negative) u² is positive.

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete