Two particles of masses m and M (M > m )are connected by a cord that passes over a mass less, friction less pulley. The tension T in the string and the acceleration a of the particles is

\(T=\frac{2mM}{(M-m)}g;a=\frac{Mm}{(M+m)}g\)
\(T=\frac{2mM}{(M+m)}g;a=\left (\frac{M-m}{(M+m)} \right )g\)
\(T=\frac{2mM}{(M+m)}g;a=\left (\frac{M-m}{(M+m)} \right )g\)
\(T=\left (\frac{mM}{(M+m)} \right )g;a=\left (\frac{2Mm}{(M+m)} \right )g\)

Solution

An explosion breaks a rock into three parts in a horizontal plane. Two of them go of fat right angles to each other.The first part of mass 1 kg moves with a speed of 12 ms_-1 and the second part of mass 2 kg moves with speed 8 ms_-1.If the third part flies off with speed 4 ms–1 then its mass is

5 kg
7 kg
17 kg
3 kg

Solution

Three blocks with masses m, 2 m and 3 m are connected by strings as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m?(g is the acceleration due to gravity)

2 mg
3 mg
6 mg
zero

Solution

From figure

F = 6 mg,

As speed is constant, acceleration a = 0

∴ 6 mg =6ma = 0, F =6 mg

∴ T = 5mg , T'= 3 mg

T"= 0

F_net on block of mass 2 m

= T – T' – 2 mg = 0

ALTERNATE :

∵ v =constant

so, a = 0, Hence, F_net= ma = 0

The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

\(\mu =\frac{2}{tan\theta }\)
μ = 2 tan θ
μ = tan θ
\(\mu =\frac{1}{tan\theta }\)

Solution

A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving up wards with an acceleration 1.0 m/s². If g = 10 ms^-2, the tension in the supporting cable is

8600 N
9680 N
11000 N
1200 N

Solution

Total mass = (60 + 940) kg = 1000 kg

Let T be the tension in the supporting cable, then

T – 1000g = 1000 × 1

⇒T = 1000 × 11 = 11000 N

A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them is µ =0.5. The distance that the box will move relative to belt before coming to rest on it taking g = 10 ms^-2, is

1.2 m
0.6 m
zero
0.4 m

Solution

Frictional force on the box f = μmg

∴ Acceleration in the box

a = μg = 5 ms^-2

v² = u² + 2as

⇒ 0 = 2² + 2 × (5) s

⇒ s = –²⁄₅ w.r.t. belt

⇒ distance = 0.4 m

body under the action of a force \(\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}\),acquires an acceleration of 1 m/s². The mass of this body must be

10 kg
20 kg
10√2 kg
2√10 kg

Solution

The coefficient of static friction, μ_s, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and mass less. (g = 10 m/s²)

0.4 kg
2.0 kg
4.0 kg
0.2 kg

Solution

m_Bg = μ_s m_Ag{∵ m_Ag =μ_s m_Ag}

⇒ m_B= μ_s m_A

or m_B = 0.2 × 2 = 0.4 kg

A block of mass m is placed on a smooth wedge of inclination q. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

mg/cos q
mg cos q
mg sin q
mg

Solution

A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal. Then its acceleration is

\(\frac{g\, sin\, \theta }{1-MR^{2}/I}\)
\(\frac{g\, sin\, \theta }{1-I/MR^{2}}\)
\(\frac{g\, sin\, \theta }{1-MR^{2}/I}\)
\(\frac{g\, sin\, \theta }{1-I/MR^{2}}\)

Solution

This is a standard formula and should be memorized.

\(a=\frac{g\, sin\, \theta }{1+\frac{I}{MR^{2}}}\)

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