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}$$
A block placed on a rough horizontal surface is pulled by a horizontal force $$F$$. Let $$f$$ be the force applied by the rough surface on the block. Plot a graph of $$f$$ versus $$F$$.
The approximate graph is shown in the diagram
The frictional force $$f$$ is shown on vertical axis and the applied force $$F$$ is shown on the horizontal axis. The portion OA of graph represents static friction which is self adjusting. In this portion, $$f=F$$.
The point $$B$$ corresponds to force of limiting friction which is the maximum value of static friction. $$C D \| O X$$ represents kinetic friction, when the body actually starts moving. The force of kinetic friction does not increase with applied force, and is slightly less than limiting friction.
Why are porcelain objects wrapped in paper or straw before packing for transportation?
Porcelain object are wrapped in paper or straw before packing to reduce the chances of damage during transportation. During transportation sudden jerks or even fall takes place, the force takes longer time to reach the porcelain objects through paper or straw for same change in momentum as $$F=\frac{\Delta p}{\Delta t}$$ and therefore, a lesser force acts on object.
Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?
When a child falls on a cement floor, her body comes to rest instantly. But $$F \times \Delta t=$$ change in momentum $$=$$ constant. As time of stopping $$\Delta t$$ decreases, therefore $$F$$ increases and hence, child feel more pain.
When she falls on a soft muddy ground in the garden the time of stopping increases and hence, $$F$$ decreases and she feels lesser pain.
A woman throws an object of mass 500 g with a speed of $$25 \mathrm{~ms}^{-1}$$.
(a) What is the impulse imparted to the object?
(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?
Mass of the object $$(m)=500 \mathrm{~g}=0.5 \mathrm{~kg}$$
Speed of the object $$(v)=25 \mathrm{~m} / \mathrm{s}$$
(a) Impulse imparted to the object = change in momentum
$$\begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =0.5(25-0)=12.5 \mathrm{~N}-\mathrm{s} \end{aligned}$$
(b) Velocity of the object after rebounding
$$\begin{aligned} & =-\frac{25}{2} \mathrm{~m} / \mathrm{s} \\ v^{\prime} & =-12.5 \mathrm{~m} / \mathrm{s} \\ \therefore \quad \text { Change in momentum } & =m\left(v^{\prime}-v\right) \\ & =0.5(-12.5-25)=-18.75 \mathrm{~N}-\mathrm{s} \end{aligned}$$