Collisions(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

COLLISIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

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
mv
- 2 mv

Discuss Question

Answer: Option D. -> - 2 mv
:
D

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

Question 2.

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

Partially elastic
Total elastic
Total inelastic
None of the above

Discuss Question

Answer: Option C. -> Total inelastic
:
C

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

Question 3.

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

Discuss Question

Answer: Option C. -> Move with the same velocity in opposite directions
:
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 4.

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

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

Discuss Question

Answer: Option C. ->

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

:
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 5.

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

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

Discuss Question

Answer: Option A. ->

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

:
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 6.

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

$\frac{1}{4}$
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{3}{4}$

Discuss Question

Answer: Option C. ->

\frac{2}{3}

:
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 7.

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

1
0
$\infty$
-1

Discuss Question

Answer: Option A. -> 1
:
A

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

Question 8.

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

$e h$
$e^{2} h$
$e^{3} h$
$e^{4} h$

Discuss Question

Answer: Option D. ->

e^{4} h

:
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 9.

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

Bob A comes to rest at B and bob B move to the left attaining a maximum height h.
Bob A and B both move to the left, each attaining a maximum height $\frac{h}{2}$ .
Bob B moves to the left and bob A moves to the right, each attaining a maximum height $\frac{h}{2}$ .
Both bobs come to rest.

Discuss Question

Answer: Option A. -> Bob A comes to rest at B and bob B move to the left attaining a maximum height h.
:
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 10.

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?

0.9 kg
1.2 kg
1.5 kg
1.8 kg

Discuss Question

Answer: Option B. -> 1.2 kg
:
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