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__Introductory Exercise 5.5__

**Ques 1: In figure m _{1} = 1 kg and m_{2} = 4 kg. Find the mass M o f the hanging block which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction less and the strings and the pulleys are light**

m

i.e., a = g tanÎ¸ â€¦(i)

N = m

For the movement of triangular block

T - N sinÎ¸ = m

For the movement of the block of mass M

Mg - T = Ma â€¦(iv)

Adding Eqs. (iii) and (iv),

Mg - N sinÎ¸ = ( m

Substituting the value of N from Eq. (ii) in the above equation

Mg - (m

=( m

i.e., M (g - Î±) = m

Substituting value of a from Eq. (i) in the above equation,

M(1 - tanÎ¸) = m

Substituting Î¸ = 30Â° , m

= 6.82 kg

Ans:

Displacement of block at time t relative to car would be

Velocity of block at time t (relative to car) will be

(b) Time (t) for the block to arrive at the original position (i.e., x = x

x

â‡’ t = 4s

Ans:

(b) x = x

In carâ€™s frame

x = x

i.e., x = x

and z = z + 10t â€¦(ii)

Velocity of the object at time t would be

and

(b) In ground frame the position of the object at time t would be given by

In ground frame

x = x

and z = z

Velocity of the object at time t would be

ans

Ans:

Normal force on object = mg

Maximum sliding friction = Î¼

= 0.3*2*10 = 6 N

Deceleration due to friction = 6/2 =3 m/s

Deceleration due to pseudo force = 5 m/s

âˆ´ Net deceleration = (3 + 5) m/s

= 8 m/s

âˆ´ Displacement of object at any time t (relative to car)

Thus, velocity of object at any time t (relative to car)

The object will stop moving relative to car when

10 - 8t = 0 i.e., t = 1.25s

âˆ´ v

Ans:

f + ma cosÎ¸ = mg sinÎ¸

or

f = mg sinÎ¸ - ma cosÎ¸

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