Q.50. Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, res

Q: What is the correct answer for JEE Main Physics Q50?

The correct answer is 9. Sol. For motion under constant acceleration, distance in $n^{\mathrm{th}}$ second is $S_n = u + \frac{a}{2}(2n-1)$. For mass 1 ($m_1 = 3.4 \mathrm{kg}$, $u_1 = 5 \mathrm{m/s}$): $104 = 5 + \frac{a_1}{2}(2 \times 5 - 1) \Rightarrow 104 = 5 + \frac{9a_1}{2} \Rightarrow 99 = \frac{9a_1}{2} \Rightarrow a_1 = 22 \mathrm{m/s}^2$. For mass 2 ($m_2 = 2.5 \mathrm{kg}$, $u_2 = 12 \mathrm{m/s}$): $129 = 12 +

Q: Q.50. Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, respectively. The distances traversed by the masses in the second are 104 m and 129 m, respectively.…

The correct answer is 9. Sol. For motion under constant acceleration, distance in $n^{\mathrm{th}}$ second is $S_n = u + \frac{a}{2}(2n-1)$. For mass 1 ($m_1 = 3.4 \mathrm{kg}$, $u_1 = 5 \mathrm{m/s}$): $104 = 5 + \frac{a_1}{2}(2 \times 5 - 1) \Rightarrow 104 = 5 + \frac{9a_1}{2} \Rightarrow 99 = \frac{9a_1}{2} \Rightarrow a_1 = 22 \mathrm{m/s}^2$. For mass 2 ($m_2 = 2.5 \mathrm{kg}$, $u_2 = 12 \mathrm{m/s}$): $129 = 12 +

Question

Q.50. Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, respectively. The distances traversed by the masses in the  second are 104 m and 129 m, respectively. The ratio of their momenta after 10 s is . The value of  is _______.

FirstInTest · Accepted Answer

Sol. For motion under constant acceleration, distance in $n^{\mathrm{th}}$ second is $S_n = u + \frac{a}{2}(2n-1)$.
For mass 1 ($m_1 = 3.4 \mathrm{kg}$, $u_1 = 5 \mathrm{m/s}$):
$104 = 5 + \frac{a_1}{2}(2 	imes 5 - 1) \Rightarrow 104 = 5 + \frac{9a_1}{2} \Rightarrow 99 = \frac{9a_1}{2} \Rightarrow a_1 = 22 \mathrm{m/s}^2$.
For mass 2 ($m_2 = 2.5 \mathrm{kg}$, $u_2 = 12 \mathrm{m/s}$):
$129 = 12 + \frac{a_2}{2}(2 	imes 5 - 1) \Rightarrow 129 = 12 + \frac{9a_2}{2} \Rightarrow 117 = \frac{9a_2}{2} \Rightarrow a_2 = 26 \mathrm{m/s}^2$.
Velocity after 10 s:
$v_1 = u_1 + a_1 t = 5 + 22 	imes 10 = 5 + 220 = 225 \mathrm{m/s}$.
$v_2 = u_2 + a_2 t = 12 + 26 	imes 10 = 12 + 260 = 272 \mathrm{m/s}$.
Momenta after 10 s:
$P_1 = m_1 v_1 = 3.4 	imes 225$.
$P_2 = m_2 v_2 = 2.5 	imes 272$.
Ratio of momenta: $\frac{P_1}{P_2} = \frac{3.4 	imes 225}{2.5 	imes 272} = \frac{765}{680} = \frac{153}{136}$.
The solution provided is $\frac{P_1}{P_2} = \frac{(5+22	imes 10)3.4}{(12+26	imes 10)(2.5)} = \frac{225 	imes 3.4}{272 	imes 2.5} = \frac{9}{8}$.
My calculation $\frac{153}{136}$ simplifies to $\frac{9 	imes 17}{8 	imes 17} = \frac{9}{8}$. So the solution is correct. Therefore, $X=9$.

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