In the above given problem if the lower thread is pulled with a jerk, what happens?
When the lower thread $$C D$$ is pulled with a jerk, the thread $$C D$$ itself break. Because pull on thread $$C D$$ is not transmitted to the thread $$A B$$ instantly.
Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in figure. Calculate $$T_1$$ and $$T_2$$ when whole system is going upwards with acceleration $$=2 \mathrm{~m} / \mathrm{s}^2$$ (use $$\mathrm{g}=9.8 \mathrm{~ms}^{-2}$$).
$$\begin{aligned} \text { Given, } m_1 & =5 \mathrm{~kg}, m_2=3 \mathrm{~kg} \\ g & =9.8 \mathrm{~m} / \mathrm{s}^2 \text { and } a=2 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$$
For the upper block
$$\begin{aligned} & & T_1-T_2-5 g & =5 a \\ \Rightarrow & & T_1-T_2 & =5(g+a) \end{aligned}$$
For the lower block
$$T_2-3 g=3 a$$
$$\Rightarrow \quad T_2=3(g+a)=3(9.8+2)=35.4 \mathrm{~N}$$
$$\begin{aligned} &\text { From Eq. (i) }\\ &\begin{aligned} T_1 & =T_2+5(g+a) \\ & =35.4+5(9.8+2)=94.4 \mathrm{~N} \end{aligned} \end{aligned}$$
Block $$A$$ of weight 100 N rests on a frictionless inclined plane of slope angle $$30^{\circ}$$. A flexible cord attached to $$A$$ passes over a frictionless pulley and is connected to block $$B$$ of weight $$w$$. Find the weight $$w$$ for which the system is in equilibrium.
In equilibrium, the force $$m g \sin \theta$$ acting on block $$A$$ parallel to the plane should be balanced by the tension in the string, i.e.,
$$\begin{aligned} m g \sin \theta & =T=F \quad [\because \text{T=F given}]\quad \text{.... (i)}\\ w & =T=F \quad \text{... (ii)} \end{aligned}$$
and for block B, where, $$w$$ is the weight of block $$B$$.
From Eqs. (i) and (ii), we get,
$$\begin{aligned} \therefore \quad W & =m g \sin \theta \\ & =100 \times \sin 30^{\circ} \quad (\because m g=100 \mathrm{~N})\\ & =100 \times \frac{1}{2} \mathrm{~N}=50 \mathrm{~N} \end{aligned}$$
A block of mass $$M$$ is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $$\mu$$ and the acceleration due to gravity is $$g$$, calculate the minimum force required to be applied by the finger to hold the block against the wall.
Given, mass of the block $$=M$$
Coefficient of friction between the block and the wall $$=\mu$$
Let a force $$F$$ be applied on the block to hold the block against the wall. The normal reaction of mass be $$N$$ and force of friction acting upward be $$f$$. In equilibrium, vertical and horizontal forces should be balanced separately.
$$\begin{array}{ll} \therefore & f=M g \quad \text{... (i)}\\ \text { and } & F=N \quad \text{... (ii)} \end{array}$$
$$\text { But force of friction }(f)=\mu \mathrm{N}$$
$$=\mu F\quad \text{[using Eq. (ii)] .... (iii)}$$
$$\begin{aligned} &\text { From Eqs. (i) and (iii), we get } \quad \mu F =M g \\ &\text { or } \quad F =\frac{M g}{\mu} \\ &\begin{aligned} \\ \end{aligned} \end{aligned}$$
A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $$=10 \mathrm{~ms}^{-2}$$)
$$\text { Given, mass of the gun }\left(m_1\right)=100 \mathrm{~kg}$$
Mass of the ball $$\left(m_2\right)=1 \mathrm{~kg}$$
Height of the cliff $$(h)=500 \mathrm{~m}$$
Horizontal distance travelled by the ball $$(x)=400 \mathrm{~m}$$
From $$h=\frac{1}{2} g t^2 \quad(\because \text { Initial velocity in downward direction is zero)}$$
$$\begin{aligned} 500 & =\frac{1}{2} \times 10 t^2 \\ t & =\sqrt{100}=10 \mathrm{~s} \end{aligned}$$
$$\text { From } \quad x=u t, u=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}$$
$$\begin{aligned} &\text { If } v \text { is recoil velocity of gun, then according to principle of conservation of linear momentum, }\\ &\begin{aligned} m_1 v & =m_2 u \\ v & =\frac{m_2 u}{m_1}=\frac{1}{100} \times 40=0.4 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$