A ball is projected vertically upwards with kinetic energy E.The kinetic energy of the ball at the highest point of its flight will be

E
E/\(\sqrt{2}\)
E/2
zero

Solution

At highest point of the trajectory velocity becomes zero and all kinetic energy changes to potential energy so at highest point, K.E. = 0

Two trains are each 50 m long moving parallel towards each other at speeds 10 m/s and 15 m/s respectively. After what time will they pass each other?

\(5\sqrt{\frac{5}{3}}\)sec
4 sec
2 sec
6 sec

Solution

Relative speed of each train with respect to each other be, v= 10 +15 = 25 m/s

Here distance covered by each train = sum of their lengths= 50 + 50 = 100 m

∴Required time=\(\frac{100}{25}\)=4sec.

A ball is dropped downwards, after 1 sec another ball is dropped downwards from the same point. What is the distance between them after 3 sec?

25 m
20 m
50 m
9.8 m

Solution

S = ut + ¹⁄₂ at² here a = g

For first body u₁=0 ⇒S₁=¹⁄₂ g × 9

For second body u₂=0 ⇒ S₂= ¹⁄₂ g × 4

So difference between them after 3 sec. = S₁ – S₂= ¹⁄₂ g × 5

If g = 10 m/sec²then S₁–S₂= 25 m.

A smooth inclined plane is inclined at an angle θ with horizontal. A body starts from rest and slides down the inclined surface.

Then the time taken by it to reach the bottom is

\(\sqrt{\left ( \frac{2h}{g} \right )}\)
\(\sqrt{\left ( \frac{2h}{g} \right )}\)
\(\sqrt{\left ( \frac{2h}{g} \right )}\)
\(sin\Theta \frac{\sqrt{(2h)}}{g}\)

Solution

A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is

3u²/g
4u²/g
6u²/g
9u²/g

Solution

The stone rises up till its vertical velocity is zero and again reached the top of the tower with a speed u(downward). The speed of the stone at the base is 3u.

A body covers 26, 28, 30, 32 meters in 10^th, 11^th, 12^th and 13^th seconds respectively. The body starts

from rest and moves with uniform velocity
from rest and moves with uniform acceleration
with an initial velocity and moves with uniform acceleration
with an initial velocity and moves with uniform velocity

Solution

The distance covered in n^th second is

From eqs. (1) and (2) we get u = 7m/sec, a=2m/sec²

∴The body starts with initial velocity u =7m/sec and moves with uniform acceleration a = 2m/sec²

A stone thrown vertically upwards with a speed of 5 m/sec attains a height H₁. Another stone thrown upwards from the same point with a speed of 10m/sec attains a height H₂.The correct relation between H₁and H₂is

H₂ = 4H₁
H₂ = 3H₁
H₁=2H₂
H₁ = H₂

Solution

From third equation of motion v²=u²+2ah

In first case initial velocity u₁ = 5 m/sec

final velocity v₁= 0, a= – g

and max. height obtained is H₁, then, H₁=\(\frac{25}{2g}\)

In second case u₂= 10 m/sec, v₂= 0, a= –g

and max. height is H₂ then, H₂=\(\frac{100}{2g}\).

It implies that H₂ = 4H₁

A particle is moving along a straight line path according to the relations 2 = at² + 2bt + cs represents the distance travelled int seconds and a,b, care constants. Then the acceleration of the particle variesas

s^–3
s^3/2
s^-3/2
s²

Solution

A train of 150 m length is going towards north direction at a speed of 10 ms^-1. A parrot flies at a speed of 5 ms^-1 to wards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to

12 s
8 s
15 s
10 s

Solution

So by figure the velocity of parrot w.r. t. train is = 5–(–10) = 15m/sec

so time taken to cross the train is

A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance s₁ in the first 10 seconds and distance s₂ in the next 10 seconds, then

s₂ = s₁
s₂ = 2 s₁
s₂ = 3 s₁
s₂ = 4 s

Solution

Let a be the constant acceleration of the particle. Then

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