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
$$\begin{aligned} & =-\frac{25}{2} \mathrm{~m} / \mathrm{s} \\ v^{\prime} & =-12.5 \mathrm{~m} / \mathrm{s} \\ \therefore \quad \text { Change in momentum } & =m\left(v^{\prime}-v\right) \\ & =0.5(-12.5-25)=-18.75 \mathrm{~N}-\mathrm{s} \end{aligned}$$
Why are mountain roads generally made winding upwards rather than going straight up?
While going up a mountain, the force of friction acting on a vehicle of mass $$m$$ is $$f=\mu R=\mu \mathrm{mg} \cos \theta$$, where $$\theta$$ is the angle of slope of the road with the horizontal. To avoid skidding force of friction ($$f$$) should be large and therefore, $$\cos \theta$$ should be large and hence, $$\theta$$ should be small.
That's why mountain roads are generally made winding upwards rather than going straight upto avoid skidding.
A mass of 2 kg is suspended with thread $$A B$$ (figure). Thread $$C D$$ of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward direction, so as to apply force on $$A B$$. Which of the threads will break and why?
The thread $$A B$$ will break earlier than the thread $$C D$$. This is because force acting on thread $$C D=$$ applied force and force acting on thread $$A B=$$ (applied force + weight of 2 kg mass). Hence, force acting on thread $$A B$$ is larger than the force acting on thread $$C D$$.
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.