JEE Questions for Physics Laws Of Motion Quiz 4 - MCQExams.com

Acceleration of block, a = force/(total mass )
= P (M +m)
∴ Force on block = mass × acceleration = Ma = MP/(M + m)
 

The change in momentum
= mv cos 60⁰ – (– m v cos 60⁰)
= 2mv cos 60⁰ = 2mv (1/2) = m v.

mg – R = ma
mg – ηmg = ma mg (1 – η) = ma, a = g (1 – η)

2 T cos 45º = mg or 2T × 1/√2 = mg
T = mg √2/2 = 5 × 10 × √2/2 = 25 √2 = 35.1 N

It is according to law of conservation of momentum i.e, m₁v₁ + m₂v₂ = 0.
 

While firing a bullet from a rifle, there will be equal back thrust on light riffle as well as heavy rifle. Due to this, the light rifle will move backwards with greater speed than heavy rifle. That is why light rifle will cause more injury.

When protons are at a given separation, the distance between them will be quite large than the size of the nucleus. Hence the nuclear forces will not be active. The electromagnetic forces are stronger than gravitational forces. Hence E > G > N.

One force is the reaction of the other, hence the tension in rope is the force applied on one end of rope =10 k N.

When a container filled with water and having a wooden block floating in it falls freely under gravity, effective weight of block becomes zero, as it is in state of weightlessness.

Let T be the tension in the string. Equation of motion of A is; T = Ma (i). If a’ is the acceleration of B, the equation of motion of B can be written as Ma’ = F – T = F – Ma ∴ a ' = (F/M) − a

Let T be tension in the string and a be the acceleration of block down the incline. On the horizontal plane, T = Ma. The equation of motion of block moving down the incline is
Ma = Mg sin θ – T or T = Mg sin θ – T
                               T = 1/2 Mg sin θ

Since the speed of block A down the plane is constant.
∴ T + F = mg sinθ
or T = mg sinθ − F = mg sinθ − μmg cos θ
T = mg [sinθ − μcos θ]newton = m [sinθ − μcos θ]kgf
T =10 [sin30^o − 0.2cos30^o ] =10 [0.5 − 0.2× 0.866] = 3.3 kgf .
Since weight of block B =Tension
T = 3.3 kgf.
∴ mass of block B = 3.3 kg

Here, m = 60 kg : μ_s = 0.5 ; μ_k = 0.4 ; a = ?
Force to be applied for just motion of block, f_s = μ_smg = 0.5 × 60 × 9.8 N  
Accelerating force = f_s – f_k =0.5 × 60 × 9.8 – 0.4 × 60 × 9.8 = 0.1 × 60 × 9.8N
Acceleration,a = (f_s – f_k)/M = 0.1 × 60 × 9.8 / 60 = 0.98 m/s²

Force acting on the block down the incline is, mg sinθ =1×10 sin 37^o = 6.018N
Force of friction acting up the incline is, F = μR = μ mg cosθ
F = 0.8 × 1 × 10 cos 37º = 6.389 N
As F > mg sin θ, the block will not slide down the incline, even when tension in the string is zero.

If T’ is tension in the string B and T = tension in string C , then in equilibrium, T
cos 45º = 100 kg and T sin 45º = T’
∴ T’ = 100 kg = 100 gN

Let T be the tension in each part of the string.
For pulley P₁, T = W1. Let ∠AP₂P₁ = 2θ
For pulley P₂, 2 T cos θ = W₂ or 2W₁ cos θ = W₂
As W₁ = W₂, cosθ = 1/2 ;θ = 60^o
∴∠AP₂P₁ = 2θ = 2 × 60^o = 120^o .

For angular displacement θ, the situation is shown in fig . Resolving mg into two rectangular components, we note that T–mg cos θ provides the required centripetal force = mv²/l
i.e. T–mg cos θ = mv²/l.

Since upward forces on the rod are equal to downward forces, hence, the resultant force on the rod is zero and there is no linear motion.Moment of forces about one end of rod =F× 20 – F × 40 + F × 60 – F ×80 = –F × 40.As it is not zero, hence a torque acts on the rod.

The weight mg of the block has two rectangular components; mg cos θ perpendicular to the plane and mg sin θ down the plane. Force of friction f = μR = μmg cos θ is also down the plane.
Therefore, F = mg sinθ + f = mg sin θ + μ mg cos θ ,But μ = tanθ

∴ F = mg sin θ + tanθ mg cos θ = 2 mg sinθ.

As the man is stationary with respect to belt, his acceleration = acceleration of belt = a = 1 m/s².The weight of man is balanced by an equal upward reaction of the belt on him. Therefore,the only force acting on the man is F = ma = 60 × 1.0 = 60 N.

According to law of conservation of momentum, when one body of mass m moving with velocity u collides head on with another body of mass m at rest, the first body comes to rest and the second body moves with velocity u. Therefore, 2 bodies each of mass m moving with velocity u collide with 6 balls in contact at rest, the motion is first conveyed to two balls, then to the next two balls and finally to the last two balls, which will start moving with velocity u.

Let velocity towards East be positive, then velocity towards west is negative. If v is the velocity of two lumps which are coalesced after collision, then following law of conservation of linear momentum, we have m₁v₁+m₂ v₂=( m₁+m₂)v or 10 ×10 –10 ×20 = 20 × v or v = – 100 /20 = – 5cm/s.

The two blocks will have the same acceleration and identical free body diagrams.Therefore, force of interaction between two blocks is zero in all cases.
 

Since the wall is smooth i.e. frictionless, the system will fall down due to its own weight, whatever be the pushing force F.