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
$$\begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =2\left(0-\frac{3}{4}\right) \\ & =-\frac{3}{2} \mathrm{~kg}-\mathrm{m} / \mathrm{s}=-1.5 \mathrm{~kg}-\mathrm{m} / \mathrm{s} \end{aligned}$$
A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?
When a person driving a car suddenly applies the brakes, the lower part of the body slower down with the car while upper part of the body continues to move forward due to inertia of motion.
If driver is not wearing seat belt, then he falls forward and his head hit against the steering wheel.
The velocity of a body of mass 2 kg as a function of $$t$$ is given by $$\mathbf{v}(t)=2 t \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}$$. Find the momentum and the force acting on it, at time $$t=2 \mathrm{~s}$$.
Given, mass of the body $,$m=2 \mathrm{~kg}$$.
Velocity of the body $$\mathbf{v}(t)=2 t \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}$$
$$\therefore$$ Velocity of the body at $$t=2 \mathrm{~s}$$
$$\mathbf{v}=2 \times 2 \hat{\mathbf{i}}+(2)^2 \hat{\mathbf{j}}=(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})$$
Momentum of the body $$(p)=m \mathbf{v}$$
$$=2(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})=(8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) \mathrm{kg}-\mathrm{m} / \mathrm{s}$$
$$\begin{aligned} \text { Acceleration of the body }(a)=\frac{d \mathbf{v}}{d t} & \\ & =\frac{d}{d t}\left(2 t \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}\right) \\ & =(2 \hat{\mathbf{i}}+2 t \hat{\mathbf{j}}) \\ \text { At } t & =2 s \\ \mathbf{a} & =(2 \hat{\mathbf{i}}+2 \times 2 \hat{\mathbf{j}}) \\ & =(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \end{aligned}$$
$$\begin{aligned} & \text { Force acting on the body }(\mathbf{F})=m \mathbf{a} \\ & \qquad \begin{aligned} & =2(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \\ & =(4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) \mathrm{N} \end{aligned} \end{aligned}$$