Consider a car moving on a straight road with a speed of 100m/s .The distance at which car can be stopped is [µ_k = 0.5]

1000 m
800 m
400 m
100 m

Solution

v² - u² = 2as or 0² - u²(-µ_kg) s

-100² = 2 × -¹⁄₂ × 10 × s

s = 1000 m

A block is kept on a friction less inclined surface with angle of inclination ‘α’ . The incline is given an acceleration ‘a’ to keep the block stationary. Then a is equal to

g cosec α
g/ tan α
g tan α
g

Solution

A particle of mass 0.3 kg subject to a force F =– kx with k =15 N/m . What will be its in itial acceleration if it is released from a point 20 cm away from the origin ?

15 m/s²
3 m/s²
10 m/s²
5 m/s²

Solution

Mass (m) = 0.3 kg ⇒ F = m.a= –15 x

a =- ¹⁵⁄_0.3x=^-150⁄₃x=50x

a =– 50 × 0.2 = 10 m/s²

The upper half of an inclined plane with inclination φ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

2 cos φ
2 sin φ
tan φ
2 tan φ

Solution

Acceleration of block while sliding down upper half =g sin φ;

retardation of block while sliding down lower half =–(g sin φ-μ g cos φ)

For the block to come to rest at the bottom, acceleration in I half = retardation in II half.

g sin φ=- (g sin φ -µ g cos φ)

⇒µ = 2 tan φ

Alternative method :According to work-energy theorem, W = ΔK = 0

(Since initial and final speeds are zero)

∴ Work done by friction + Work done by gravity= 0

i.e., -(µ mg cos φ)^ι⁄₂ + mg ι sin φ = 0

or ^µ⁄₂cos φ=sin φ or µ = 2 tan φ

A smooth block is released at rest on a 45° incline and then slides a distance ‘d’. The time taken to slide is ‘n’ times as much to slide on rough incline than on a smooth incline.The coefficient of friction is

\(\mu _{k}=\sqrt{1-\frac{1}{n^{2}}}\)
\(\mu _{k}=1-\frac{1}{n^{2}}\)
\(\mu _{s}=\sqrt{1-\frac{1}{n^{2}}}\)
\(\mu _{s}=1-\frac{1}{n^{2}}\)

Solution

In the given figure, the pulley is assumed mass less and friction less. If the friction force on the object of mass m is f,then its acceleration in terms of the force F will be equal to

(F - f)/m
(^F⁄₂-f)/m
F/m
None of these

Solution

A ball mass m falls vertically to the ground from a height h₁ and rebounds to a height h₂. The change in momentum of the ball of striking the ground is

m √2 g(h₁ + h₂)
\(n\sqrt{2g(m_{1}+m_{2})}\)
mg(h₁ - h₂)
\(m(\sqrt{2gh_{1}}-\sqrt{2gh_{2}})\)

Solution

An open topped rail road car of mass M has an initial velocity v₀ along a straight horizontal friction less track. It suddenly starts raising at time t= 0. The raindrops fall vertically with velocity u and add a mass m kg/sec of water. The velocity of car after t second will be (assuming that it is not completely filled with water)

v₀+m^u⁄_M
\(\frac{mv_{0}}{M+mt}\)
\(\frac{Mv_{0}+ut}{M+ut}\)
\(\frac{Mv_{0}+ut}{M+ut}\)

Solution

The rain drops falling vertically with velocity u do not affect the momentum along the horizontal track. A vector has no component in a perpendicular direction Rain drops add to the mass of the car

Mass added in t sec = (mt) kg

Momentum is conserved along horizontal track.

Initial mass of car = M

Initial velocity of car = v₀

Final velocity of (car + water) = v

Mass of(car + water) after time t = (M + mt)

∴ final momentum = initial momentum

(M + mt)v = Mv₀

∴\(v=\frac{Mv_{0}}{(M+mt)}\)

Block A of weight 100 kg rests on a block B and is tied with horizontal string to the wall at C. Block B is of 200 kg. The coefficient of friction between A and Bis 0.25 and that between B and surface is ¹⁄₃. The horizontal force F necessary to move the block B should be (g =10 m/s²)

1050 N
1450 N
1050 N
1250 N

Solution

A ball of mass 400 gm is dropped from a height of 5 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 newton so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is (g = 10 m/s²)

0.12 s
0.08 s
0.04 s
12 s

Solution

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