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**Q 1. A particle (m = 1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P (height 1 m), the kinetic energy of the particle (in J) is (Figure drawn is schematic and not to scale; take g = 10 m/s ^{2}) ______. [2020]Ans: **10

Given that m = 1 kg, g = 10 m/s

According to law of conservation of energy, we have

Energy at point A = Energy at point P

⇒ (Potential energy + Kinetic energy)

⇒ mgh

Since, K

2mg + 0 = mg + K

⇒ K

= 1 x 10 = 10 J

Given that

Capacity of elevator, C = 10 persons

Average mass of each person, m_{p} = 68 kg

Mass of elevator, m_{e} = 920 kg

Speed of elevator, v = 3 m/s

Frictional force, f = 600 N

Force acting on motor is given by,

F = m_{total} x g + f

= ( 10 x 68 + 920) x 10 + 6000

=16000 + 6000

= 22000

So, power delivered by motor is P = F x v

= 22000 x 3 = 66000 W

Therefore, power delivered by the motor to the elevator must be at least 66000 W.**Q 3. A body A, of mass m = 0.1 kg has an initial velocity of It collides elastically with another body, B of the same mass which has an initial velocity of After collision, A moves with a velocity The energy of B after collision is written as (x/10) J. The value of x is ______. ****Ans: **1**Solution:**

Given that

Since, collision is elastic so by the conservation of linear momentum

Initial momentum = Final momentum

Kinetic energy of B after collision is

Therefore, x = 1**Q 4. A particle of mass m is dropped from a height h above the ground. At the same time, another particle of the same mass is thrown vertically upwards from the ground with a speed of If they collide head-on completely inelastically, the time for the combined mass to reach the ground, in units of is [2020]****Ans: **(4)**Solution:**

Given that

m_{A} = m_{B} = m, v_{0,A} = v_{0,B} =

Time for collision is given by

velocity before collision is

v_{i,A} = gt, v_{i,B} = V_{0,B} - gt

⇒ v_{i,A} = -v_{i,B}

From conservation of momentum, we have

m_{A}v_{i.A} + m_{B}v_{i,B} = m_{A}v_{f,A} + m_{B}v_{f,B}

So, height from ground is given by**Q 5. Consider a force The work done (in J) by this force in moving a particle from point A(1, 0) to B(0, 1) along the line segment given in the fiigure is (all quantities are in SI units) [2020]****(1) 2 ****(2) 1/2****(3) 1****(4) 3/2****Ans: **(3)**Solution:**

Given that

Work done by the force is given by**Q 6. Two particles of equal mass m have respective initial velocities They collide completely inelastically. The energy lost in the process is [2020]****(1) 1/3 mμ ^{2}**

Given that

Since collision is inelastic only momentum is conserved. So,

Change in kinetic energy is given by

ΔK = K

Given that initial speed u, angle of projection from ground θ = π/3

Speed of particle at height point is given by

Applying conservation of momentum, we have

Time of flight after collision is given by

Range after collision is given by

(1)

(2)

(3)

(4)

Ans:

The maximum speed of block is at mean position F = kx (equilibrium)

According to work energy theorem,

Work done by external force + work done by spring = Change in kinetic energy

**(1) 5(2) 2(3) 4(4) 3Ans:** (3)

From the linear momentum, we have

Now, according to the question,

**Q 10. A force acts on a 2 kg object so that its position is given as a function of time as x = 3t ^{2} + 5. What is the work done by this force in first 5 s? (2019)(1) 850 J(2) 950 J(3) 875 J(4) 900 JAns: **(4)

The position of object as a function of time is given by

x = 3t

We know that

Differentiating w.r.t. Eq. (1) becomes

Therefore, work done by the force = change in K.E.

(1)

(2)

(3) 0

(4)

Ans:

According to the Newton second law

F = ma

Work done is

(1) 20 m

(2) 30 m

(3) 40 m

(4) 10 m

Ans:

Bullet will collide with piece of wood at

Velocity of piece of wood u = 0 + 10 × 1 = 10 m/s

Velocity of bullet v = 100 – 10 = 90 m/s

From conservation of linear momentum,

−(0.02) (1v) + (0.02) (9v) = (0.05)

⇒ 150 = 5v

⇒ v = 30 m/s

Maximum height reached by body is

At time of collision system is h distance below top of tower

Therefore, height above tower = 45 – 5 = 40 m

(2) 12 J

(3) 10 J

(4) 15 J

Ans:

According to work-energy theorem

(1) 8 cm

(2) 4 cm

(3) 80 cm

(4) 40 cm

* None of the options is correct

Initial compression is mg = kx

Velocity of 1 kg block just before it collides with 3 kg block

Applying momentum conversation just before and just after collision, we have

Now, applying work-energy theorem,

(1) 2m

(2) 3.5m

(3) 1.5m

(4) 4m

Ans:

Applying law of conservation of momentum before collision,

mu

⇒ mu

Since, α-particle is scattered in backward thus, losing 64% of initial energy.

Now, coefficient of restitution

thus

(1)

(2)

(3) Zero

(4)

Ans:

Solution:

‘ – ‘ sign of force implies attractive force So,**Q 17. In a collinear collision, a particle with an initial speed ν _{0} strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is: (2018)(1) (2) (3) (4) Ans:** (2)

By conservation of linear momentum,

So relative velocity :

Hence the Solution is Option (2)

(1) Tα R

(2)

(3) Tα R

(4) Tα R

Ans:

**Q 19. A body of mass m starts moving from rest along x-axis so that its velocity varies as v = a√s where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first 't' seconds after the start of the motion is: (2018)(1) (8) ma ^{4}t^{2}(2) (1/4) ma^{4}t^{2}(3) (4) ma^{4}t^{2}(4) (1/8) ma^{4}t^{2}Ans: **(4)

(1) 3 × 10

Ans:

(1) 10

(2) 10

(3) 10

(4) 10

Ans:

(1) 9 J

(2) 18 J

(3) 4.5 J

(4) 22 J

Ans:

(1) 3h

(2) ∞

(3) 5/3h

(4) 8/3h

Ans:

Ans:

A: 2.45 x 10

B: 6.45 x 10

C: 9.89 x 10

D: 12.89 x 10

Ans:

0.2 x 3.8 x 107 x m = 10 x g x 1 x 1000

m =

= 1.289 x 10

= 12.89 x 10

(1) 12000 J

(2) –12000 J

(3) –4500 J

(4) –9300 J

Ans:

(1) log v(t) against t

(2) v(t) against t

(3) log v(t) against 1/t

(4) log v(t) against 1/t

Ans:

logV vs 1/t will be a st. line curve.

(1) M n R2 t2

(2) 1/2 Mn

(3) Mn

(4) MnR

Ans:

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