Mass $$m_1$$ moves on a slope making an angle $$\theta$$ with the horizontal and is attached to mass $$m_2$$ by a string passing over a frictionless pulley as shown in figure. The coefficient of friction between $$m_1$$ and the sloping surface is $$\mu$$. Which of the following statements are true?
In figure a body $$A$$ of mass $$m$$ slides on plane inclined at angle $$\theta_1$$ to the horizontal and $$\mu$$ is the coefficient of friction between $$A$$ and the plane. $$A$$ is connected by a light string passing over a frictionless pulley to another body $$B$$, also of mass $$m$$, sliding on a frictionless plane inclined at an angle $$\theta_2$$ to the horizontal. Which of the following statements are true?
Two billiard balls $$A$$ and $$B$$, each of mass 50 g and moving in opposite directions with speed of $$5 \mathrm{~m} \mathrm{~s}^{-1}$$ each, collide and rebound with the same speed. If the collision lasts for $$10^{-3} \mathrm{~s}$$, which of the following statements are true?
A body of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is
A girl riding a bicycle along a straight road with a speed of $$5 \mathrm{~ms}^{-1}$$ throws a stone of mass 0.5 kg which has a speed of $$15 \mathrm{~ms}^{-1}$$ with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg . Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?
Given, total mass of girl, bicycle and stone $$=m_1=(50+0.5) \mathrm{kg}=50.5 \mathrm{~kg}$$.
Velocity of bicycle $$u_1=5 \mathrm{~m} / \mathrm{s}$$, Mass of stone $$m_2=0.5 \mathrm{~kg}$$
Velocity of stone $$u_2=15 \mathrm{~m} / \mathrm{s}$$, Mass of girl and bicycle $$m=50 \mathrm{~kg}$$
Yes, the speed of the bicycle changes after the stone is thrown.
Let after throwing the stone the speed of bicycle be $$v \mathrm{~m} / \mathrm{s}$$.
According to law of conservation of linear momentum,
$$\begin{aligned} m_1 u_1 & =m_2 u_2+m v \\ 50.5 \times 5 & =0.5 \times 15+50 \times v \\ 252.5-7.5 & =50 \mathrm{v} \\ \text{or}\quad v & =\frac{245.0}{50} \\ v & =4.9 \mathrm{~m} / \mathrm{s} \\ \text { Change in speed } & =5-4.9=0.1 \mathrm{~m} / \mathrm{s} . \end{aligned}$$