Free Objective Test 01 Practice Test - 11th and 12th

Question 1

A body of mass 'm' collides against a wall with a velocity 'v' and retraces its path with the same speed. The change in momentum is (take initial direction of velocity as positive)

Zero

2 mv

- 2 mv

SOLUTION

Solution : D

Change in momentum = $m {\to v}_{2} - m {\to v}_{1} = - m v - m v = - 2 m v$

Question 2

A shell initially at rest explodes into two pieces of equal mass, then the two pieces will

Be at rest

Move with different velocities in different directions

Move with the same velocity in opposite directions

Move with the same velocity in same direction

SOLUTION

Solution : C

According to law of conservation of linear momentum both pieces should possess equal momentum after explosion. As their masses are equal therefore they will possess equal speed in opposite direction.

Question 3

A particle of mass 'm' moving eastward with a speed 'v' collides with another particle of the same mass 'm' and moving northward with the same speed 'v'. The two particles coalesce on collision. The new particle of mass '2'm will move in the north-easterly direction with a velocity

$\frac{v}{2}$

$\frac{v}{\sqrt{2}}$

SOLUTION

Solution : C

Initial momentum of the system

${\to P}_{i} = m v^i + m v^j$

$| {\to P}_{i} |$ = $\sqrt{2} m v$

Final momentum of the system = 2mV

By the law of conservation of momentum

$\sqrt{2} m v = 2 m V \Rightarrow V = \frac{v}{\sqrt{2}}$

Question 4

When two bodies stick together after collision, the collision is said to be

Partially elastic

Total elastic

Total inelastic

None of the above

SOLUTION

Solution : C

By definition of Perfectly Inelastic Collision, the two colliding bodies stick together after collision

Question 5

A body of mass 2 kg moving with a velocity of 3 m/sec collides head on with a body of mass 1 kg moving in opposite direction with a velocity of 4 m/sec. After collision, two bodies stick together and move with a common velocity which in m/sec is equal to

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{3}{4}$

SOLUTION

Solution : C

$m_{1} v_{1} - m_{2} v_{2} = (m_{1} + m_{2}) v$

$\Rightarrow (2 \times 3) - (1 \times 4) = (2 + 1) v \Rightarrow v = \frac{2}{3} m / s$

Question 6

The coefficient of restitution 'e' for a perfectly elastic collision is

$\infty$

-1

SOLUTION

Solution : A

No energy is lost in perfectly elastic collision. Therefore e = 1

Question 7

A sphere of mass 'm' moving with a constant velocity 'u' hits another stationary sphere of the same mass. If 'e' is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be

$\frac{1 - e}{1 + e}$

$\frac{1 + e}{1 - e}$

$\frac{e + 1}{e - 1}$

$\frac{e + 1}{e - 1} t^{2}$

SOLUTION

Solution : A

Let the final velocities of the two masses be $v_{1}$ and $v_{2}$ By definition, $e = \frac{v_{2} - v_{1}}{0 - 4}$
Conservation of linear momentum, $m (u) + 0 = m v_{1} + m v_{2}$
$\Rightarrow v_{1} + v_{2} = 4$
$∴ \frac{v_{1} + v_{2}}{v_{2} - v_{1}} = \frac{4}{- e 4}$
$\frac{v_{1} + v_{2}}{v_{1} - v_{2}} = \frac{1}{e}$
$∴ \frac{v_{1}}{v_{2}} = \frac{1 + e}{1 - e}$
$\frac{v_{2}}{v_{1}} = \frac{1 - e}{1 + e}$

Question 8

The bob A of a pendulum released from a height 'h' hits head-on another bob B of the same mass of an identical pendulum initially at rest. What is the result of this collision? Assume the collision to be elastic

A. Bob A comes to rest at B and bob B move to the left attaining a maximum height h.

B. Bob A and B both move to the left, each attaining a maximum height

\frac{h}{2}

C. Bob B moves to the left and bob A moves to the right, each attaining a maximum height

\frac{h}{2}

D. Both bobs come to rest.

SOLUTION

Solution : A

Suppose the bob A acquires a velocity v on reaching the bob B. In a head-on elastic collision between two bodies of the same mass, the velocities are exchanged after the collision. Hence the bob A will come to rest at the lowermost position (occupied by B before collision) and the bob B will move to the left attaining a maximum height h. hence the correct choice is (a).

Question 9

A steel ball falls from a height h on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is

e h

e^{2} h

e^{3} h

e^{4} h

SOLUTION

Solution : D

The velocity attained after a fall through a height h is given by $v^{2} = 2 g h$
Thus $h \propto v^{2}$ The velocity after first rebound is ev. Therefore, the height attained after first rebound = $e^{2} h$ . Velocity after second rebound is $e^{2} v$ . Hence the height attained after second rebounds is $e^{4} h$ . Thus the correct choice is (d).

Question 10

Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in Fig. Ball 1 moves with a velocity $v _{1}$ towards balls 2 and 3. All collisions are assumed to be elastic. If M <m, the number of collisions between the balls will be

A. One

B. Two

C. Three

D. Four

SOLUTION

Solution : B

The first collision will be between balls 1 and 2. Since both have the same mass, after the collision ball 1 will come to rest and ball 2 will move with speed $v_{1}$ . This ball will collide with the stationary ball 3. After this second collision, let $v_{2}$ and $v_{3}$ be the speeds of balls 2 and 3 respectively. Since the collisions are elastic, $v_{2}$ and $v_{3}$ are given by (see Sec. 9)
$v_{2} = (\frac{m - M}{m + M}) v_{1}$ (i)
and $v_{3} = (\frac{2 m}{m + M}) v_{1}$ (ii)
If M < m, it follows from (i) and (ii) that v₂ < v₃ and both have the same direction.
Therefore, ball 2 cannot collide with ball 3 again. Hence there are only two collisions.
Thus the correct choice is (b).

Question 11

A ball of mass 'm' moving horizontally at a speed 'v' collides with the bob of a simple pendulum at rest. The mass of the bob is also' 'm, if the collision is perfectly elastic, the bob of the pendulum will rise to a height of

\frac{v^{2}}{g}

\frac{v^{2}}{2 g}

\frac{v^{2}}{4 g}

\frac{v^{2}}{8 g}

SOLUTION

Solution : B

In an elastic collision between two bodies of the same mass with one of them initially at rest, the moving body is brought to rest and the other moves in the same direction with the same speed. Thus the ball will come to rest and the bob of the pendulum acquires a speed v. At this speed, it will rise to height h given by $h = \frac{v^{2}}{2 g}$ . Hence the correct choice is (b).

Question 12

A ball P of mass 2 kg undergoes an elastic collision with another ball Q initially kept at rest. After the collision, ball P continues to move in its original direction with a speed one-fourth of its original speed. What is the mass of ball Q?

A. 0.9 kg

B. 1.2 kg

C. 1.5 kg

D. 1.8 kg

SOLUTION

Solution : B

Since the collision is elastic, both momentum and kinetic energy are conserved. If m is mass of ball Q and v’ its speed after the collision, the law of conservation of momentum gives
$2 v = 2 \times \frac{v}{4} + m v^{^{'}}$
Where v is the original speed of ball P. thus
$m v^{^{'}} = \frac{3 v}{2} o r v^{^{'}} = \frac{3 v}{2 m}$ (i)
The law of conservation of energy gives
$\frac{1}{2} \times 2 \times v^{2} = \frac{1}{2} \times 2 \times {(\frac{v}{4})}^{2} + \frac{1}{2} m v^{^{'} 2}$
or $m v^{^{'} 2} = \frac{15 v^{2}}{8}$ (ii)
Using (i) in (ii) we get m = 1.2 kg. Hence the correct choice is (b).

Question 13

A neutron moving at a speed 'v' undergoes a head-on elastic collision with a nucleus of mass number 'A' at rest. The ratio of the kinetic energies of the neutron after and before collision is

{(\frac{A - 1}{A + 1})}^{2}

{(\frac{A + 1}{A - 1})}^{2}

{(\frac{A}{A + 1})}^{2}

{(\frac{A}{A - 1})}^{2}

SOLUTION

Solution : A

Mass of neutron $(m_{1}) = 1 u n i t$ . Mass of nucleus $(m_{2}) = A u n i t s$ .
The velocity of the neutron after the collision is
$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u = (\frac{1 - A}{1 + A}) u$
KE of neutron after collision = $\frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} \times 1 \times {(\frac{1 - A}{1 + A})}^{2} u^{2}$
KE of neutron before collision = $\frac{1}{2} m u^{2} = \frac{1}{2} \times 1 \times u^{2} = \frac{1}{2} u^{2}$ .
Their ratio is ${(\frac{1 - A}{1 + A})}^{2}$ , which is choice (a)

Question 14

A body of mass 5 kg is moving along the x-axis with a velocity 2 $m s^{- 1}$ . Another body of mass 10 kg is moving along the y-axis with a velocity $\sqrt{3} m s^{- 1}$ . If they collide at the origin and stick together, then the final velocity of the combined mass is

\sqrt{3} m s^{- 1}

(\sqrt{3} + 1) m s^{- 1}

\frac{4}{3} m s^{- 1}

N o n e

SOLUTION

Solution : C

Momentum of 5 kg mass $(p_{1}) = 5 \times 2 = 10 k g m s^{- 1}$ along the x-axis. Momentum of 10 kg mass $(p_{2}) = 10 \sqrt{3} k g m s^{- 1}$ along the y-axis. These two momenta are perpendicular to each other.
Therefore, the resultant initial momentum is
$P = \sqrt{p_{1}^{2} + p_{2}^{2}} = \sqrt{(10)^{2} + (10 \sqrt{3})^{2}} = 20 k g m s^{- 1}$
If v $m s^{- 1}$ is the velocity of the combined mass, then the final momentum = (10 + 5) v = 15 v kg $m s^{- 1}$
Now, from the principle of conservation of momentum, we have 15 v = 20 or $v = \frac{4}{3} m s^{- 1}$ , which is choice (c).

Question 15

A body of mass 'm' moving with a velocity 'v' in the x-direction collides with a body of mass 'M' moving with a velocity 'V' in the y-direction. They stick together during a collision. Then

A. The magnitude of the momentum of the composite body is

\sqrt{(m v)^{2} + (M V)^{2}}

B. The composite body moves in a direction making a angle

θ = t a n^{- 1} (\frac{M V}{m v})

with the x-axis.

C. The loss of kinetic energy as a result of collision is

\frac{1}{2} \frac{M m}{(M + m)} (V^{2} + v^{2})

D. All the above choices are correct.

SOLUTION

Solution : D

Refer to Figure Here p = mv and P = MV.

The resultant of p and P is
$p_{r} = \sqrt{p^{2} + P^{2}} = \sqrt{(m v)^{2} + (M V)^{2}}$
Which is choice (a), the angle which the resultant momentum $p_{r}$ subtends with the x-axis is given by
$t a n θ = \frac{P}{p} = \frac{M V}{m v}$ , which is choice (b).
Loss of KE = $(\frac{1}{2} m v^{2} + \frac{1}{2} M V^{2}) - [\frac{1}{2} \frac{m^{2} v^{2} + M^{2} V^{2}}{(M + m)}]$
= $\frac{1}{2} \frac{M m}{(M + m)} (V^{2} + v^{2})$ , which is choice (c).
Hence the correct choice is (d).

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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