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**MULTIPLE CHOICE QUESTIONS **

**Q.1. A ball is travelling with uniform translatory motion. This means that****(a) It is at rest.****(b) The path can be a straight line or circular and the ball travels with uniform speed.****(c) All parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.****(d) The centre of the ball moves with constant velocity and the ball spins about its centre uniformly.****Ans. **(c)**Solution.**

When a body moves in such a way that the linear distance covered by each particle of the body is same during the motion, then the motion is said to be translatory or translation motion.

Translatory motion can be, again of two types viz., curvilinear (shown in fig. (a)) or rectilinear (shown in fig. (b)), accordingly as the paths of every constituent particles are similarly curved or straight line paths. Here it is important that the body does not change its orientation. Here we can also define it further in uniform and non-uniform translatory motion. Here figure

(b) is uniformly translatory motion.

**Q.2. ****A metre scale is moving with uniform velocity. This implies(a) The force acting on the scale is zero, but a torque about the centre of mass can act on the scale.(b) The force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.(c) The total force acting on it need not be zero but the torque on it is zero.(d) Neither the force nor the torque need to be zero.**

As the meter, the scale is moving with uniform velocity.

∴ No change in its velocity i.e., acceleration of it zero by Newton’s second law

∴ Hence net or resultant force must act on body zero.

So torque must be zero.

Hence, for uniform motion force and torque, both must be zero.So, It verifies the option (b).

(a) zero

(b)

(c)

(d)

m = 150g = 0.15kg

Change in momentum momentum-Initial momentum

Hence, this verifies the option(c).

(b) 0.75 kg m s

(c) 1.5 kg m s

(d) 14 kg m s

Ans.

Solution.

By previous solution,

= 1.5 kg ms

So, this verifies the option (c).

(a) Conservation of energy.

(b) Newton’s first law only.

(c) Newton’s second law only.

(d) Both Newton’s second and third law.

(i) By Newton’s second law

As in law of conservation of momentum is zero.

i.e., = 0

= 0

⇒

(ii) By Newton’s third law action force is equal to reaction force in magnitude but in opposite direction.

∴ ( = 0)

So proves the law of conservation of momentum and verifies the option (d).

(a) Frictional force along westward.

(b) Muscle force along southward.

(c) Frictional force along south-west.

(d) Muscle force along south-west.

As we know rate of change of momentum is force So in this case direction of force acting on player will be the same as the direction of change in momentum. Now to understand this in better let’s look the image below

shows player’s northward and westward directions respectively. Now shows south direction sois towards south-west which means force on player will act on this direction.

Therefore, option (c) is correct.

The force acting on the body at t = 2 seconds is

(a) 136 N

(b) 134 N

(c) 158 N

(d) 68 N

We have given in question x(t) = pt + qt

We have x(t) = pt + qt

So x(t) = 3t + 4t

Now first derivative of above equation

Therefore, option (a) is correct.

(b) 10 s

(c) 2 s

(d) 15 s

We have to find the time at which the final velocity along the y axis which means x-component will be zero. We have now X-component = v

Therefore, option (b) is correct.

(a) mv/2 eastward and is exerted by the car engine.

(b) mv/2 eastward and is due to the friction on the tyres exerted by the road.

(c) more than mv/2 eastward exerted due to the engine and overcomes the friction of the road.

(d) mv/2 exerted by the engine.

Car start from rest mean initial velocity u = 0, also and t = 2s mass is given as m.

As we know

Now force exerted on the car i.e. here this force is due to engine of car so whatever this force is actually internal force in another words due to friction force the car moves in eastward direction.

(a) The force at t = (1/8) s on the particle is –16π

(b) The particle is acted upon by on impulse of magnitude 4π

(c) The particle is not acted upon by any force.

(d) The particle is not acted upon by a constant force.

(e) There is no impulse acting on the particle.

We have given mass = m, x(t) = 0 for t < 0

Now let’s look for time interval

x(t) = A sin 4πt

Also we know,

1

2

Then,

Here, this force is obviously time dependent so force is not constant hence option (d) is correct.

Now let’s look to option (a) i.e. At

So from this we can say option (a) also correct.

Now for option (b) i.e. between t = 0 s and t = (1/4) s, as we know impulse is change in momentum or so at t = (1/4) s impulse will be

here we have to keep in mind that F(t) varies from t=0 to maximum 1/8 s, so F (1/4) will be replaced by F (1/8) also from above we have found

so option(b) is also correct.

Option (c) is absolutely wrong as we have above found that force is acting.

Option (e) is also wrong because we have already calculated the impulse.

(a) The bodies will move together if F = 0.25 mg.

(b) The body A will slip with respect to B if F = 0.5 mg.

(c) The bodies will move together if F = 0.5 mg.

(d) The bodies will be at rest if F = 0.1 mg.

(e) The maximum value of F for which the two bodies will move together is 0.45 mg.

We have given Let acceleration in body A and B is ‘a’.

Body A will move along with body B by force F till the force of friction between surface of A and B is larger or equal to zero.

Now taking system A+B then acceleration will be

So force on A

If F

N = Reaction force by B on A

[N

From (i)

so adding this with equation (ii) we can say F = 0.45mg ...(iii)

F = 0.45mg Newton is the maximum force on B. so that A and B can move together. So option (e) is correct.

Both bodies can move together if F is less than or equal to 0.45mg Newton.

So options (a) and (b) are also correct and rejects the option (c) as 0.5mg>0.45mg.

For option (d): Minimum force which can move A and B together

Given force in option (d) 0.1mg Newton <0.25mg Newton. So body A and B will not move i.e.

Bodies A and B will remain in rest hence option (d) is also correct.

(a) If m

(b) If m

(c) If m

(d) If m

Let’s consider a case in which normal reaction i.e.

N = m

Now from figure taking whole as a system then

when m

Putting f in this equation

from this option (a) is totally wrong while option (b) is correct.

Now if body m

Hence option (d) is correct but option (c) is wrong.

(c) For A to move up the plane, θ

(d) B will always slide down with constant speed.

Condition | Free body diagrams |

Equation | Tension and acceleration |

T - m_{1}g sin α = m_{1}a | |

m_{2}a = m_{2}g sin β - T |

In this problem first we have to decide the direction of motion. Let block A moves up the plane friction force on A will be downward (along the plane) as shown.**Q.14. ****Two billiard balls A and B, each of mass 50g and moving in opposite directions with speed of 5m s ^{–1} each, collide and rebound with the same speed. If the collision lasts for 10^{–3} s, which of the following statements are true?**

(a) The impulse imparted to each ball is 0.25 kg m s^{–1} and the force on each ball is 250 N.

(b) The impulse imparted to each ball is 0.25 kg m s^{–1} and the force exerted on each ball is 25 × 10^{–5} N.

(c) The impulse imparted to each ball is 0.5 Ns.

(d) The impulse and the force on each ball are equal in magnitude and opposite in direction.

According to the problem, balls are identical.

Initial velocity (u) = u_{1} = u_{2} = 5 m/s

Final velocity (v) = v_{1} = v_{2} = -5 m/s

Time duration of collision = 10^{-3}s

Change in linear momentum = m(v - u)

Impulse and force are opposite in directions.**Q.15. ****A body of mass 10kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is(a) 1 m s ^{–2} at an angle of tan^{-1} (4/3) w.r.t. 6N force.(b) 0.2 m s^{–2} at an angle of tan^{-1} (4/3) w.r.t. 6N force.(c) 1 m s^{–2} at an angle of tan^{-1} (3/4) w.r.t. 8N force.(d) 0.2 m s^{–2} at an angle of tan^{-1} (3/4) w.r.t. 8N force.**

Recall the concept of resultant of two vectors, when they are perpendicular

As they are perpendicular, cos 90° = 0

So resultant will be

As shown in the diagram

According to the problem, mass = m = 10 kg

F

Let θ

Let θ

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.1.**** A girl riding a bicycle along a straight road with a speed of 5 m s ^{–1} throws a stone of mass 0.5 kg which has a speed of 15 m s^{–1} with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?**

Also we have given u

As we know by law of conservation of momentum initial momentum (girl, bicycle, stone) = final momentum (cycle and girl) + stone

Therefore, the speed of cycle and girl decreased by

So W' = R

= mg - ma

= m(g - a)

Apparent weight due to reaction force by the lift on weighing scale will be

Reading of weighing scale = R/g = 50/10 = 5 kg

We have given mass of body m=2 kg, initial velocity (v1) at t=0 is zero. Now in the interval t ≥ 0 to t ≤ 4 x-t graph is straight line that mean velocity of body will remain constant.

And when t ≤ 4 the slope of graph is zero so velocity will also be zero i.e. v

Now or Impulse=change in momentum

Impulse at t = 0

Impulse at t = 4

So from above observation impulse at t = 0 increased by + 1.50 kg ms

So momentum at 2 sec p (2) = mv (2)

Hence momentum and force at 2 sec will be and respectively.

Ans.

Now when a small force F

After force f

In diagram A is equivalent to limiting frictional force and a B is equivalent to kinetic frictional force.

When objects are packed in paper or straw etc. the objects takes more time to stop or change velocity during jerks (due to breaks, or uneven road) so acceleration (v - u)/t decreased. So the force on objects will be smaller and objects become safer.

On cemented hard floor the time to stop after fall on it is very-very small. But when she/he falls on soft ground of garden she/he sinks in ground and takes more time to stop hence smaller force or pain acts on her/him.

(a) What is the impulse imparted to the object?

(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?

u = 0, v = 25 m/s

Hence, the or force is opposite to the initial velocity of ball.**Q.10. ****Why are mountain roads generally made winding upwards rather than going straight up?****Ans. **On an inclined plane force of friction on a body going upward is f_{s} = μN cos θ where θ is angle of inclination of a plane with horizontal if θ is small, the force of friction is high and there is a less chance of skidding. The road straight up would have a larger angle and smaller would be the value of friction, hence more are the chances of skidding.

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