PHYSICS – DYNAMICS
Physics – Dynamics
OBJECTIVE TYPE QUESTION
Question – 1.
A bullet of mass m is fired with a velocity v in a large target of mass M. The final velocity V would be :
(A) Mv/(m + M) (B) mv/(m + M) (C) (m + M)v/m (D) (m + M)V/M
Solution : –
We have, as momentum is conserved,
mv = (m + M)V
Or, V = mv/(m + M) (B) [Ans.]
Question – 2.
A force of 100 N acts on a huge mass of 100 kg for 0.1 second. The change in momentum would be :
(A) 10 kgms–1 (B) 10.5 kgms–1 (C) 110 kgms–1 (D) 100 kgms–1
Solution : –
We have, Impulse = Change in momentum
= Force × time
= 100 × 0.1
= 10 kgms–1 (A) [Ans.]
Question – 3.
A ball of mass m impinges upon a stone with velocity v and returns with the same velocity. The change in the momentum of the ball is given by :
(A) mv (B) – mv (C) – 2mv (D) 2mv
Solution : –
We have, Momentum before impact = mv
Momentum after impact = – mv
As, direction of velocity is reversed,
Change in momentum = mv – (– mv)
= 2mv (D) [Ans.]
Question – 4.
Two spheres of masses 3 kg and 4 kg are tied to a string passing over a frictionless pulley. Taking g to be 9.8 ms–2, the relative acceleration of the system would be :
(A) 1.4 ms–2 (B) 3 ms–2 (C) 4 ms–2 (D) 7 ms–2
Solution : –
We have, Acceleration = [(m2 – m1)/(m2 + m1)]g
= [(4 – 3)/(4 + 3)]×9.8
= 1.4 ms–2 (A) [Ans.]
Question – 5.
A person of mass 50 kg descends in a lift with an acceleration 2 ms–2. The cable of lift suddenly breaks down. The weight of the person inside the lift would be :
(A) 50 kg (B) 52 kg (C) 48 kg (D) zero
Solution : –
We have, the apparent weight of the person,
N = m(g – a ), where, a = 2 ms–2.
When cable breaks, a = g
Therefore, N = m(g – g)
= 0 (D) [Ans.]
Question – 6.
The mass of a body in a lift at rest is m. When the lift ascends with a uniform acceleration a, the effective mass of the body would be :
(A) m(1 + g/a) (B) m(1 – g/a) (C) m(1 + a/g) (D) m(1 – a/g).
Solution : –
We have, When the lift ascends with uniform acceleration a, the apparent weight of the body,
W = m(g + a)
Let effective mass in the lift be m1,
Then m1 = W/g
= {m(g + a)}/g
= m(1 + a/g) (C) [Ans.]
Question – 7.
A rifle shoots a bullet of weight 50 g with a velocity of 60 m/s. If the velocity of recoil of rifle is 2 m/s, the mass of the rifle would be :
(A) 1.0 kg (B) 1.5 kg (C) 2.0 kg (D) 3.0 kg
Solution : –
We have, if mass of the rifle is M,
Momentum of the rifle = Momentum of the bullet
Or, M × 2 = 50 × 60
= 3000
Or, M = 3000/2
= 1500 g
= 1.5 kg (B) [Ans.]
Question – 8.
An object of weight W rests on a horizontal plane. If angle of friction be θ, then minimum force required to displace the object along the horizontal plane would be :
(A) W sin θ (B) W cos θ (C) W tan θ (D) W cot θ
Solution : –
We have, minimum force required to displace an object of weight W on a horizontal plane is equal to the force of friction = μR = μmg = μW
Here, μ = tan θ,
Therefore, force of friction = W tan θ (C) [Ans.]
Question – 9.
If μ be the coefficient of friction between the tyres and the road then the least stopping distance for a car of mass m moving with velocity v is given by :
(A) v2/2μg (B) v/2μg (C) v2/μg (D) 2v/μ2g
Solution : –
We have, using v2 = u2 + 2as,
where, u = v, a = – μg, v = 0, s = ?
We get, 0 = v2 + 2 (– μg)s
Or, s = v2/2μg (A) [Ans.]
Question – 10.
A satellite sweeps stationary interplanetary dust in a force free space at the rate of dM/dt = βν. The acceleration of the satellite would be :
(A) βν2/M (B) Mβ/ν2 (C) –βν2/M (D) – Mβ/ν
Solution : –
We have, Thrust on the satellite,
F = –ν.dM/dt
= – ν(βν)
= – βν2
Acceleration = Force/Mass
= – βν2/M (C) [Ans.]
Question – 11.
A bird of 2.5 kg is carried in a cage of 1.5 kg inside it. When the bird starts flying inside the cage, the weight of bird and cage assembly is given by :
(A) 1 kg (B) 1.5 kg (C) 2.5 kg (D) 4 kg
Solution : –
We have, Weight of bird and cage = (2.5 + 1.5)
= 4 kg (D) [Ans.]
Question – 12.
A force of (2i + 3j + 4k) N acts on a body for 19 seconds and displaces (3i + 4j + 5k) m.The power consumed would be :
(A) 19 W (B) 38 W (C) 2 W (D) none of these
Solution : –
We have, Power = W/t
W = Force . Displacement
= (2i + 3j + 4k).(3i + 4j + 5k)
= 38 J
Power = W/t
= 38/19
= 2 W (C) [Ans.]
Question – 13.
A force varies with time and traces a sine curve. The peak value of force is F0. The average force would be :
(A) Zero (B) F0 √2 (C) √F0 (D) F0
Solution : –
We have, Average force = [∫Fdt]/t
= (Area enclosed by F.t .graph)/t
= 0 (A) [Ans.]
Question – 14.
The coefficient of friction between road and tyre of a car increases linearly with distance, x. The car attains maximum possible acceleration starting from rest. The kinetic energy, E of the car depends on x as :
(A) E α1/x (B) E α x (C) E α x2 (D) E α x1/2
Solution : –
We have, μ = Kx
Therefore, a = μg
= Kxg
Using v = u + at
Or, v = 0 + Kxgt
K.E. = 1/2 mv2
= 1/2 mK2x2g2t2
Therefore, E α x2 (C) [Ans.]
Question – 15.
A cricket player catches a ball of mass 150 g in 0.1 s, moving with a velocity of 20 m/s. The force experienced by him would be :
(A) 15 N (B) 30 N (C) 3 N (D) 0.3 N
Solution : –
We have, F = rate of change of momentum
= dp/dt
= (0.150×20)/0.1
= 30 N (B) [Ans.]
Question – 16.
A man is standing on a lift with a bucket full of water. The bucket has hole in its bottom and water is flowing out with a rate of Ro. If the lift starts moving up and down with same acceleration and rate of flow of water are Ru and Rd, the :
(A) Ro > Rd > Ru (B) Rd > Ru > Ro (C) Ru > Ro > Rd (D) Ru > Rd > Ro
Solution : –
We have, as rate of flow of water is proportional to effective force pressure on water.
When the lift moves upward, Ru α m(g + a)
When the lift moves downward, Rd α m(g – a)
When the lift is stationary, Ro α mg
Therefore, Ru > Ro > Rd (C) [Ans.]
Question – 17.
A force f1 accelerate a particle from rest to a velocity of v. Another force f2 decelerates it to rest, then :
(A) f1 and f2 must be zero (C) f1 and f2 must be equal
(B) f1 and f2 may be equal (D) f1 must not be equal to f2
Solution : –
We have, according to condition of motion two forces must act in opposite direction and they may be equal, not necessarily must be equal. (B) [Ans.]
Question – 18.
A man is standing in the middle of a block of smooth ice. He can’t get to the shore by using Newton’s :
(A) third law of motion (C) first and second law of motion
(B) first law of motion (D) second law of motion
Solution : –
Man can’t get to the shore using Newton’s first and second law of motion. (C) [Ans.]
Question – 19.
A horse pulls a cart, the force that helps the horse to move forward is the force exerted by :
(A) ground on the horse (C) cart on the ground
(B) horse on the ground (D) ground on the cart
Solution : –
We have, the horse pulls the cart on the due to a component of force of reaction of ground on the horse acting horizontally. (A) [Ans.]
Question – 20.
A particle is moving with a velocity of v1. It is given an impulse and its velocity becomes v2. If its mass be m then the magnitude of impulse would be :
(A) m(v1 + v2) (B) m(v2 – v1) (C) m(v1 – v2) (D) zero
Solution : –
We have, Impulse = Change in momentum
= m(v2 – v1) (B) [Ans.]
Question – 21.
The fuel burns in a rocket at the rate of 1 kg/s. the fuel is ejected from the rocket with a velocity of 60 km/s. The force exerted on the rocket would be :
(A) 60 N (B) 600 N (C) 6,000 N (D) 60,000 N
Solution : –
We have, Thrust = v (dm/dt)
= (60×1000)×1
= 60,000 N (D) [Ans.]
Question – 22.
Mass of a rocket is M, rate of ejection of gases is r and velocity of gases with respect to rocket is u, then the acceleration of the rocket would be :
(A) ru/(M + rt) (B) ru/(M – rt) (C) ru/M (D) (M – rt)/ru.
Solution : –
We have, Initial mass of rocket = M;
dm/dt = r ;
relative velocity of gasses w.r.t. rocket = v;
Therefore acceleration of the rocket, a = F/m
= u(dm/dt)/[M – {(dm/dt)×t}]
= ur/(M – rt) (B) [Ans.]
Question – 23.
A dish of mass 10 g is kept horizontally in space by firing bullets of mass 5 g each at the rate of 10 per second. The bullets rebounds with the same speed, the velocity of the bullets would be :
(A) 98 cm/s (B) 49 cm/s (C) 9.8 cm/s (D) 4.9 cm/s.
Solution : –
We have, if velocity of the dish be v, then
Total change in momentum per second = Weight of the dish
Or, 5×10×2×v = 10×980
Or, v = 98 cm/s (A) [Ans.]
Question – 24.
A rocket of mass 10000 kg is projected vertically upward s. The exhausted gases has velocity 1000 m/s w.r.t. the rocket vertically downward. The maximum rate of burning of fuel to just lift the rocket upward against gravity would be :
(A) 196 kg/s (B) 147 kg/s (C) 98 kg/s (D) 49 kg/s
Solution : –
We have, Mg = udm/dt
Or, dm/dt = Mg/u
= (10000×9.8)/1000
= 98 kg/s (C) [Ans.]
Question – 25.
A car starts from rest to cover a distance s. The coefficient of friction between road and tyres is μ. The minimum time in which the car would cover the distance is proportional to :
(A) 1/√μ (B) √μ (C) μ (D) 1/μ
Solution : –
We have, Using s = ut + 1/2at2
As, u = 0, s = 1/2at2
= 1/2(μg)t2
Therefore, t = √(2s/μg)
Thus t α 1/√μ (A) [Ans.]
Question – 26.
Two point masses m and M are at a distance L apart. The distance of the centre of mass of the system from m is :
(A) L(M/m) (B) L(m/M) (C) LM/(m + M) (D) Lm/(M + m)
Solution : –
We have, if distance of centre of mass from the point mass m be x
Therefore, mx = M(L – x)
Or, x = ML/(m + M) (C) [Ans.]
Question – 27.
The net force acting on the system of particle is zero. Which of the following may vary :
(A) Kinetic energy of the system (C)Velocity of centre of mass
(B) Position of centre of mass (D) momentum of the system.
Solution : –
We have, net force zero.
Therefore, motion has to be uniform.
Hence, only position of centre of mass will vary. (B) [Ans.]
Question – 28.
A man hangs from a rope attached to a hot-air balloon. The mass of the man is greater than the mass of the balloon and its content. The system is stationary in air. The man now climbs upto the balloon by the rope, the centre of mass of man and balloon will :
(A) Remain stationary (B) is variable (C) moves up (D) move down.
Solution : –
We have, when man climbs, the force action and reaction starts working and no external force is involved. Therefore centre of mass of the system remain stationary. (A) [Ans.]
Question – 29.
All the particles of a body are at a distance r from the origin. The distance of the centre of mass of the body from the origin would be :
(A) equal to r (B) more than r (C) less than r (D) none of these.
Solution : –
We have, the centre of mass would be at the same distance r from the origin. (A) [Ans.]
Question – 30.
A rocket of initial mass m, moving with velocity v, discharges gas of mean density ρ and effective area a. The minimum value of v of fuel gas, enabling the rocket to rise vertically above it would be :
(A) √(ρga/m) (B) √(ρg/ma) (C) √(mg/ρa) (D)√(ρ/mga)
Solution : –
We have, dm/dt = avρ
And F = vdm/dt
= av2ρ
Also, F = mg
Therefore, v = √(mg/ρa) (C) [Ans.]
Question – 31.
A force, F→ = 6i – 8j + 10k is applied to a mass and acceleration produced is 1 m/s2. The mass of the body would be :
(A) 2√10 kg (B) 10 kg (C)20 kg (D) 10√2 kg.
Solution : –
We have, Force, F→ = 6i – 8j + 10k, and acceleration, a = 1 m/s2
Therefore, Mass = |F→|/a
= |6i – 8j + 10k|/1
= √(36 + 64 + 100)
= 10√2 (D) [Ans.]
Question – 32.
A bullet is fired from a gun with a force, F = 600 – 2 × 105 t, where F is in Newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet would be :
(A) 0 N-s (B) 0.9 N-s (C) 1.8 N-s (D) 3.6 N-s.
Solution : –
We have, when F = 0, 600 – 2 × 105 t = 0
Or, t = 600/(2 × 105)
= 3 × 10–3 s.
Now, Impulse,I = 0∫tFdt
= 0∫t(600 – 2 ×105 t) dt
= 600t – 2 × 105 × (t/2)
= 600 × 3 × 10–3 – 105 × (3 × 10–3)2
= 1.8 – 0.9
= 0.9 (B) [Ans.]
Question – 33.
A mass of 1 kg is suspended by a thin string. It is lifted up with an acceleration of 4.9 m/s2, and lowered with an acceleration of 4.9 m/s2. The ratio of the tension in two cases would be :
(A) 3 : 1 (B) 2 : 1 (C) 1 : 2 (D) 1 : 3.
Solution : –
We have, Upward acceleration, ma = T1 – mg
Therefore, T1 = m(g + a)
Downward acceleration, ma = mg – T2
Or, T2 = m(g – a)
Therefore, T2/T1 = (g + a)/(g – a)
= (9.8 + 4.9)/(9.8 – 4.9)
= 3 : 1 (A) [Ans.]
Question – 34.
A block is pushed on a horizontal surface momentarily with an initial velocity V. If μ is the coefficient of sliding friction between the block and the surface. Block would come to rest after time :
(A) g μ/V (B) g/μV (C) V/gμ (D) V/g.
Solution : –
We have, u = V; final velocity = 0
Using v = u + at
0 = V – at
Or, t = V/a
Now, F = μR
= μmg
Therefore, retardation = μg
Hence t = V/a
= V/gμ (C) [Ans.]
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