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,
$$\begin{aligned} m_1 u_1 & =m_2 u_2+m v \\ 50.5 \times 5 & =0.5 \times 15+50 \times v \\ 252.5-7.5 & =50 \mathrm{v} \\ \text{or}\quad v & =\frac{245.0}{50} \\ v & =4.9 \mathrm{~m} / \mathrm{s} \\ \text { Change in speed } & =5-4.9=0.1 \mathrm{~m} / \mathrm{s} . \end{aligned}$$
A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of $$9 \mathrm{~ms}^{-2}$$, what would be the reading of the weighing scale? $$\left(g=10 \mathrm{~ms}^{-2}\right)$$
When a lift descends with a downward acceleration a the apparent weight of a body of mass $$m$$ is given by
$$w^{\prime}=R=m(g-a)$$
Mass of the person $$m=50 \mathrm{~kg}$$
Descending acceleration $$a=9 \mathrm{~m} / \mathrm{s}^2$$
Acceleration due to gravity $$g=10 \mathrm{~m} / \mathrm{s}^2$$
Apparent weight of the person,
$$\begin{aligned} R & =m(g-a) \\ & =50(10-9) \\ & =50 \mathrm{~N} \\ \therefore \quad \text { Reading of the weighing scale } & =\frac{R}{g}=\frac{50}{10}=5 \mathrm{~kg} . \end{aligned}$$
The position-time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 0 s and t = 4 s.
Given, mass of the body $$(m)=2 \mathrm{~kg}$$
From the position-time graph, the body is at $$x=0$$ when $$t=0$$, i.e., body is at rest.
$$\therefore$$ Impulse at $$t=0, s=0$$, is zero
From $$t=0 \mathrm{~s}$$ to $$t=4 \mathrm{~s}$$, the position-time graph is a straight line, which shows that body moves with uniform velocity.
Beyond $$t=4 \mathrm{~s}$$, the graph is a straight line parallel to time axis, i.e., body is at rest $$(v=0)$$.
Velocity of the body = slope of position-time graph
$$=\tan \theta=\frac{3}{4} \mathrm{~m} / \mathrm{s}$$
Impulse (at $$t=4 \mathrm{~s})=$$ change in momentum
$$\begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =2\left(0-\frac{3}{4}\right) \\ & =-\frac{3}{2} \mathrm{~kg}-\mathrm{m} / \mathrm{s}=-1.5 \mathrm{~kg}-\mathrm{m} / \mathrm{s} \end{aligned}$$