Center of Mass and Collision Questions — NEET Physics

Q1

A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a height of 10 m . The impulse imparted to the ball during its collision with the ground is (Take

g=9.8 \mathrm{~m} / \mathrm{s}^2

)

A 0

B 84 NS

C 21 NS

D 7 NS

Correct Answer

Option C

Solution

\begin{aligned} & v_1=\sqrt{2 g h_1} \\ & =\sqrt{2 \times 9.8 \times 40} \\ & v_1=\sqrt{784}=28 \mathrm{~m} \mathrm{~s}^{-1} \\ & \text{ and } v_2=\sqrt{2 g h_2}=\sqrt{2 \times 9.8 \times 10} \\ & =\sqrt{196}=14 \mathrm{~m} \mathrm{~s}^{-1} \\ & \text{ Impulse }=\Delta \vec{p}=m\left(\vec{v}_f-\vec{v}_i\right)=m\left(\vec{v}_2-\vec{v}_1\right) \\ & =\frac{1}{2}(14-(-28)) \\ & =21 \mathrm{NS} \end{aligned}

Q2

Two bodies

A

and

B

of same mass undergo completely inelastic one dimensional collision. The body

A

moves with velocity

v_1

while body

B

is at rest before collision. The velocity of the system after collision is

v_2

. The ratio

v_1: v_2

is

A

1: 2

B

2: 1

C

4: 1

D

1: 4

Correct Answer

Option B

Solution

In a completely inelastic collision, the two bodies stick together and move with a common final velocity.

Here, before the collision, body

A

is moving with a velocity

v_1

and body

B

is at rest.

The conservation of momentum must hold true because no external forces are acting on the system.

The momentum before the collision is only due to body

A

since body

B

is at rest. Therefore, the total initial momentum

p_{\text{initial}}

of the system is given by:

p_{\text{initial}} = m_A v_1 + m_B v_0 = mv_1 + 0 = mv_1

where:

m_A

and

m_B

are the masses of bodies

A

and

B

respectively,

m

is the mass of each body,

v_1

is the velocity of body

A

,

v_0 = 0

is the velocity of body

B

because it is initially at rest. Since they undergo a completely inelastic collision, bodies

A

and

B

stick together after the collision and hence move with a common velocity

v_2

. The total mass of the combined system post-collision is

m_A + m_B = m + m = 2m

. The momentum after the collision is given by:

p_{\text{final}} = (m_A + m_B)v_2 = 2m v_2

Applying the conservation of momentum (since no external force implies momentum is conserved), we equate

p_{\text{initial}}

and

p_{\text{final}}

:

mv_1 = 2m v_2

Dividing through by

m

yields:

v_1 = 2v_2

Thus, solving for the ratio

\frac{v_1}{v_2}

gives:

\frac{v_1}{v_2} = 2

Therefore, the ratio of

v_1

to

v_2

is

2:1

, making the correct answer: Option B:

2:1

Q3

A bullet of mass

m

hits a block of mass

M

elastically. The transfer of energy is the maximum, when :

A

M=m

B

M=2 m

C

M < < m

D

M >> m

Correct Answer

Option A

Solution

In elastic collision maximum energy is transfer when

M=m

Q4

Two particles A and B initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of A is v and speed of B is 3v, the speed of centre of mass is :

A 2v

B zero

C v

D 4v

Correct Answer

Option B

Solution

Final velocity of centre of mass = Initial velocity of centre of mass = 0 because net external force on system is zero.

Q5

A

1 \mathrm{~kg}

object strikes a wall with velocity

1 \mathrm{~m} \mathrm{~s}^{-1}

at an angle of

60^{\circ}

with the wall and reflects at the same angle. If it remains in contact with wall for

0.1 \mathrm{~s}

, then the force exerted on the wall is :

A

30 \sqrt{3} \mathrm{~N}

B Zero

C

10 \sqrt{3} \mathrm{~N}

D

20 \sqrt{3} \mathrm{~N}

Correct Answer

Option C

Solution

F=\left|\frac{\Delta \vec{p}}{\Delta t}\right|=\frac{2 m v \sin \theta}{\mathrm{t}}=\frac{2(1)(1) \sin 60^{\circ}}{0.1}=10 \sqrt{3} \mathrm{~N}

Q6

The distance covered by a body of mass 5 g having linear momentum 0.3 kg m/s in 5 s is :

A 0.3 m

B 300 m

C 30 m

D 3 m

Correct Answer

Option B

Solution

P = mv

0.3 = 5 \times {10^{ - 3}}\,v

v = {{300} \over 2}

v = 60

m/s

S = v \times t

= 60 \times 5

= 300

m

Q7

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is :

A v

B

\sqrt2

v

C 2

\sqrt2

v

D 3

\sqrt2

v

Correct Answer

Option C

Solution

Momentum of the system would remain conserved.

Initial momentum = 0 Final momentum should also be zero.

Let masses be 2m, 2m and m Momentum along x-direction = 2mv

\widehat i

Momentum along y-direction = 2mv

\widehat j

Net momentum =

\sqrt {{{(2mv)}^2} + {{(2mv)}^2}} = \sqrt 2 \,.\,2

mv Now, 2

\sqrt2

mv = mv' v' = 2

\sqrt2

v

Q8

Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is :

A

{{10} \over 3}

m

B

{{20} \over 3}

m

C 10 m

D 5 m

Correct Answer

Option B

Solution

{X_{cm}} = {{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}

= {{10 \times 0 + 20 \times 10} \over {10 + 20}}

= {{200} \over {30}}

= {{20} \over 3}

m

Q9

A ball of mass 0.15 kg is dropped from a height 10m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s 2 ) nearly :

A 1.4 kg m/s

B 0 kg m/s

C 4.2 kg m/s

D 2.1 kg m/s

Correct Answer

Option C

Solution

To calculate the impulse imparted to the ball, we need to look at the change in momentum during the collision with the ground.

The impulse $I$ can be determined by the following relationship:

I = \Delta p

Where: $\Delta p$ is the change in momentum.

Momentum $p$ is defined as the product of the mass $m$ of an object and its velocity $v$ .

Since the ball rebounds to the same height, its speed just before it hits the ground and just after it leaves the ground will be the same (ignoring air resistance), although the direction of the velocity will change.

Let's calculate the speed of the ball just before the impact.

The ball drops from a height $h$ with initial velocity $u = 0$ m/s.

Using the kinematic equation for constant acceleration under gravity $g$ , we have:

v^2 = u^2 + 2gh

Plugging in the values for $u = 0$ , $g = 10$ m/s 2 , and $h = 10$ m, we get:

v^2 = 0^2 + 2 \cdot 10 \cdot 10

v^2 = 200

v = \sqrt{200}

v = 10\sqrt{2}

m/s Since the ball bounces back to the same height, its speed on the way up just after the collision will be $10\sqrt{2}$ m/s but in the opposite direction.

The change in velocity $\Delta v$ is:

\Delta v = v_{\text{final}} - v_{\text{initial}}

\Delta v = 10\sqrt{2} - (-10\sqrt{2})

\Delta v = 20\sqrt{2}

m/s Given that the mass $m$ is 0.15 kg, we can now calculate $\Delta p$ :

\Delta p = m \Delta v

\Delta p = 0.15 \cdot 20\sqrt{2}

\Delta p = 3\sqrt{2}

\Delta p \approx 3 \cdot 1.414

\Delta p \approx 4.242

kg m/s The magnitude of the impulse imparted to the ball is therefore approximately 4.242 kg m/s, which most closely matches: Option C: 4.2 kg m/s.

Q10

Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of :

A 50 cm

B 67 cm

C 80 cm

D 33 cm

Correct Answer

Option B

Solution

Let's assume both the particles are lying on X-axis.

Let the 5 kg particle is at the origin.

Then the X-coordinate of the centre of mass of the system is,

x_{cm}=\frac{m_{{}_{1}}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}

=

\frac{5\times 0+100\times 10}{5+10}

= 66.66 cm

\simeq

67 cm