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

Question

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 5th 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 _____. [2026]

Smart Achievers · Accepted Answer

(9)

$104 = 5 + \frac{a_{1}}{2} (10 - 1) \Rightarrow a_{1} = 22 {m/s}^{2}$

$129 = 12 + \frac{a_{2}}{2} (10 - 1)$

$\Rightarrow a_{2} = \frac{117 \times 2}{9} = 13 \times 2 = 26 {m/s}^{2}$

$v_{1} = 5 + 22 \times 10 = 225 m/s$

$v_{2} = 12 + 26 \times 10 = 272 m/s$

$Momentum, P = m v$

$\frac{P_{1}}{P_{2}} = \frac{m_{1} v_{1}}{m_{2} v_{2}} = \frac{3.4 \times 225}{2.5 \times 272} = \frac{765}{680} = \frac{9}{8} \Rightarrow x = 9$

Solution

Q.
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 5th 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 _____. [2026]

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Solution

Q. 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 5th second are 104 m and 129 m, respectively. The ratio of their momenta after 10 s is x8. The value of x is _____. [2026]

Ans. (9) 104=5+a12(10-1)⇒a1=22 m/s2 129=12+a22(10-1) ⇒a2=117×29=13×2=26 m/s2 v1=5+22×10=225 m/s v2=12+26×10=272 m/s Momentum, P=mv P1P2=m1v1m2v2=3.4×2252.5×272=765680=98⇒x=9

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Q.
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 5th 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 _____. [2026]