Laws of Motion MCQ Questions & Answers in Basic Physics | Physics

Total mass $$ = \left( {60 + 940} \right)kg = 1000\,kg$$
Let $$T$$ be the tension in the supporting cable, then
$$\eqalign{ & T - 1000g = 1000 \times 1 \cr & \Rightarrow T = 1000 \times 11 = 11000\,N \cr} $$

$$\eqalign{ & {T_2}\sin \theta + {T_1}\sin {60^ \circ } = W \cr & {T_1}\cos {60^ \circ } = {T_2}\cos \theta \cr & {T_2}\sin \theta + {T_2}\cos \theta .\tan {60^ \circ } = W \cr & {T_2}\left( {\sin \theta + \sqrt 3 \cos \theta } \right) = W \cr & 2{T_2}\left( {\sin {{30}^ \circ }\sin \theta + \cos {{30}^ \circ }\cos \theta } \right) = W \cr & \cos \left( {{{30}^ \circ } - \theta } \right)\,{\text{is maximum}} \cr & {\text{If}}\,{30^ \circ } - \theta = 0;\theta = {30^ \circ } \cr} $$

$$\eqalign{ & {\text{Given,}}\,{m_A} = 4\,kg \cr & {m_B} = 2\,kg \Rightarrow {m_C} = 1\,kg \cr} $$

So, total mass $$\left( M \right) = 4 + 2 + 1 = 7\,kg$$
Now, $$F = Ma \Rightarrow 14 = 7a \Rightarrow a = 2\,m/{s^2}$$
$$FBD$$  of block $$A,$$

$$\eqalign{ & F - F' = 4a \cr & \Rightarrow F' = 14 - 4 \times 2 \cr & \Rightarrow F' = 6\,N \cr} $$

$$\eqalign{ & {\text{Given: }}{m_1} = 5kg;{\text{ }}{m_2} = 10kg;{\text{ }}\mu = {\text{ }}0.15{\text{ }} \cr & {\text{For }}{m_1},{m_1}g - T = {m_1}a{\text{ }} \cr & \Rightarrow 50 - T = 5 \times a \cr & {\text{and,}}\,T - 0.15{\text{ }}\left( {m + {\text{ }}10} \right){\text{ }}g = \left( {10 + m} \right)a{\text{ }} \cr & {\text{For rest }}a = 0{\text{ }} \cr & {\text{or, }}50 = 0.15\left( {m + {\text{ }}10} \right)10 \cr} $$

$$\eqalign{ & \Rightarrow 5 = \frac{3}{{20}}\left( {m + {\text{ }}10} \right) \cr & \frac{{100}}{3} = m + {\text{ }}10\,\,\,\,\therefore m = 23.3kg;\,\,{\text{close to option}}\left( {\text{B}} \right) \cr} $$

In case (i) we have
$$\eqalign{ & {T_1} - \left( {1 \times g} \right) = 1 \times 4.9 \cr & \Rightarrow {T_1} = 9.8 + 4.9 = 14.7\,N \cr} $$
In case (ii),
$$\eqalign{ & 1 \times g - {T_2} = 1 \times 4.9 \cr & \Rightarrow {T_2} = 9.8 - 4.9 = 4.9\,N \cr & \therefore \frac{{{T_1}}}{{{T_2}}} = \frac{{14.7}}{{4.9}} = \frac{3}{1} \cr} $$

Common acceleration of system is
$$\eqalign{ & a = \frac{F}{{{m_1} + {m_2} + {m_3}}} \cr & \therefore {\text{Force}}\,{\text{on}}\,{m_3}\,{\text{is}}\,{F_3} = {m_3} \times a = \frac{{{m_3}F}}{{{m_1} + {m_2} + {m_3}}} \cr} $$

Let the velocity of the ball just when it leaves the hand is $$u$$ then applying,
$${v^2} - {u^2} = 2as$$   for upward journey
$$ \Rightarrow - {u^2} = 2\left( { - 10} \right) \times 2 \Rightarrow {u^2} = 40$$
Again applying $${v^2} - {u^2} = 2as$$   for the upward journey of the ball, when the ball is in the hands of the thrower,
$${v^2} - {u^2} = 2as$$
$$\eqalign{ & \Rightarrow 40 - 0 = 2\left( a \right)0.2 \Rightarrow a = 100\,m/{s^2} \cr & \therefore F = ma = 0.2 \times 100 = 20\,N \cr & \Rightarrow N - mg = 20 \Rightarrow N = 20 + 2 = 22\,N \cr} $$

For the wedge,

$$\eqalign{ & N'\sin \theta = F \cr & \therefore N' = \frac{F}{{\sin \theta }}.{\text{As}}\,\sin \theta < 1,\,{\text{and}}\,{\text{so}}\,N' > F \cr} $$

$$a = \frac{{{m_2}}}{{{m_1} + {m_2}}}g = \frac{3}{{7 + 3}} = 3\,m/{s^2}$$

Acceleration of the system
$$\eqalign{ & a = \frac{{F - 4g - 2g}}{{4 + 2}} = \frac{{120 - 40 - 20}}{6} \cr & = 10\,m{s^{ - 2}} \cr} $$

From figure
$$\eqalign{ & T - 2g = 2a \cr & T = 2\left( {a + g} \right) = 2\left( {10 + 10} \right) \cr & = 40\,N \cr} $$