What is x in momentum ratio p₁/p₂=x/8 for these two masses after 10s?

x=9. a₁=22, a₂=26 m/s². v₁=225, v₂=272 m/s. p₁/p₂=765/680=9/8.

Two masses accelerated from initial speeds distances in 5th second given — ratio of momenta after 10s is x/8 find x

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PhysicsKinematics

QInteger TypeMechanics

Two masses 3.4 kg and 2.5 kg start with speeds 5 m/s and 12 m/s. Distances in 5th second are 104 m and 129 m. Ratio of momenta after 10 s is $x/8$. Find $x$.

✅ Correct Answer

x = 9

✓

Solution

Find accelerations using s_n formula

$s_n=u+a\!\left(n-\dfrac{1}{2}\right)$: for $n=5$: $s_5=u+\dfrac{9a}{2}$

$a_1=\dfrac{2(104-5)}{9}=22\text{ m/s}^2,\quad a_2=\dfrac{2(129-12)}{9}=26\text{ m/s}^2$

Velocities after 10s

$v_1=5+22\times10=225\text{ m/s},\quad v_2=12+26\times10=272\text{ m/s}$

Momentum ratio

$\dfrac{p_1}{p_2}=\dfrac{3.4\times225}{2.5\times272}=\dfrac{765}{680}=\dfrac{9}{8}\Rightarrow x=\boxed{9}$

📘 Distance in nth Second: s_n = u + a(n−½)

$s_n=u+a(2n-1)/2$ gives distance in only the $n$th second. Solve for $a$ using given $s_5$. Then $v=u+at$ after $t=10$s. Momentum $p=mv$.

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Important Concepts

s_n Formulas_n=u+a(2n-1)/2. For n=5: s₅=u+9a/2. This is the distance traversed ONLY during the 5th second.

Acceleration Calculationa₁: 104=5+9a₁/2 → 9a₁/2=99 → a₁=22 m/s². a₂: 129=12+9a₂/2 → 9a₂/2=117 → a₂=26 m/s².

Velocity After 10sv₁=5+22×10=225 m/s. v₂=12+26×10=272 m/s.

Momentum Ratiop₁/p₂=3.4×225/(2.5×272)=765/680=9/8. x=9.

FAQs

What does 'distance in 5th second' mean?

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Distance covered ONLY during the 5th second (from t=4s to t=5s), not total distance from start.

Derive the s_n formula.

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s_n=½a(n²)-½a(n-1)²=½a(2n-1)=a(n-½) (from origin). With initial u: s_n=u+a(n-½)=u+a(2n-1)/2.

What is a₁?

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104=5+9a₁/2 → 9a₁/2=99 → a₁=22 m/s².

What is a₂?

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129=12+9a₂/2 → 9a₂/2=117 → a₂=26 m/s².

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026 Section B.

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