Question 102: Motion in 1D

Question

Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of $5 \mathrm{~m} / \mathrm{s}$ and $12 \mathrm{~m} / \mathrm{s}$, respectively. The distances traversed by the masses in the $5^{	ext {th }}$ second are 104 m and 129 m , respectively. The ratio of their momenta after 10 s is $\frac{x}{8}$. The value of $x$ is $\_\_\_\_$。

Accepted Answer

The displacement of an object during a $\mathrm{n}^{	ext {th }}$ second is the difference between the displacement in n seconds and the displacement in ( $\mathrm{n}-1$ ) seconds.
$$ S_n=S(n)-S(n-1) $$
Using the second equation of motion ( $S=u t+\frac{1}{2} a t^2$ ):
$$ S_n=\left[u n+\frac{1}{2} a n^2ight]-\left[u(n-1)+\frac{1}{2} a(n-1)^2ight] $$
$$\Rightarrow $$ $$ S_n=u+\frac{a}{2}(2 n-1) $$
For mass $1\left(\mathrm{~m}_1=3.4 \mathrm{~kg}, \mathrm{u}_1=5 \mathrm{~m} / \mathrm{s}, \mathrm{S}_5=104 \mathrm{~m}ight)$ :
$$ 104=5+\frac{a_1}{2}(2 	imes 5-1) $$
$$\Rightarrow $$ $99=\frac{a_1}{2}(9)$
$$\Rightarrow $$ $$ 11=\frac{a_1}{2} \Rightarrow a_1=22 \mathrm{~m} / \mathrm{s}^2 $$
For Mass $2\left(\mathrm{~m}_2=2.5 \mathrm{~kg}, \mathrm{u}_2=12 \mathrm{~m} / \mathrm{s}, \mathrm{S}_5=129 \mathrm{~m}ight)$ :
$$ 129=12+\frac{a_2}{2}(2 	imes 5-1) $$
$$\Rightarrow $$ $117=\frac{a_2}{2}(9)$
$$\Rightarrow $$ $$ 13=\frac{a_2}{2} \Rightarrow a_2=26 \mathrm{~m} / \mathrm{s}^2 $$
Using the first equation of motion $(\mathrm{v}=\mathrm{u}+\mathrm{at})$ for $\mathrm{t}=10 \mathrm{~s}$ :
Velocity of mass 1 is,
$$ \mathrm{v}_1=5+(22 	imes 10)=5+220=225 \mathrm{~m} / \mathrm{s} $$
Velocity of mass 2 is,
$$ v_2=12+(26 	imes 10)=12+260=272 \mathrm{~m} / \mathrm{s} $$
Momentum is defined as the product of mass and velocity ( $\mathrm{p}=\mathrm{mv}$ ).
The momentum of mass 1 is $\mathrm{p}_1=\mathrm{m}_1 \mathrm{v}_1=3.4 	imes 225=765 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}$
The momentum of mass 2 is $\mathrm{p}_2=\mathrm{m}_2 \mathrm{v}_2=2.5 	imes 272=680 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}$
So, the ratio of their momenta is :
$$ \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{765}{680} $$
$$\Rightarrow $$ $$ \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{9}{8}=\frac{\mathrm{x}}{8} \Rightarrow \mathrm{x}=9 $$
Therefore, the value of x is 9 .

Motion in 1D

Integer / Numeric Answer Type