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Kinematics

JEE Main level, numerical answer, medium

118 of 308

Question 118 of 308

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 $5^\text{th}$ second are $104$ m and $129$ m, respectively. The ratio of their momenta after $10$ s is $\dfrac{x}{8}$ . The value of $x$ is ________.

Show answer and solution

Answer: 9

The distance traversed in the

n^{\text{th}}

second is given by

S_n = u + \dfrac{a}{2}(2n - 1)

.

For the first mass (

m_1 = 3.4

kg,

u_1 = 5

m/s):

104 = 5 + \dfrac{a_1}{2}(2 \times 5 - 1)

99 = \dfrac{9a_1}{2} \Rightarrow a_1 = 22

m/s

^2

For the second mass (

m_2 = 2.5

kg,

u_2 = 12

m/s):

129 = 12 + \dfrac{a_2}{2}(2 \times 5 - 1)

117 = \dfrac{9a_2}{2} \Rightarrow a_2 = 26

m/s

^2

Velocities of the masses after

10

v_1 = u_1 + a_1 t = 5 + 22 \times 10 = 225

m/s

v_2 = u_2 + a_2 t = 12 + 26 \times 10 = 272

m/s

Momenta of the masses after

10

p_1 = m_1 v_1 = 3.4 \times 225 = 765

kg m/s

p_2 = m_2 v_2 = 2.5 \times 272 = 680

kg m/s

Ratio of their momenta:

\dfrac{p_1}{p_2} = \dfrac{765}{680} = \dfrac{9}{8}

Comparing with

\dfrac{x}{8}

, we get

x = 9

.

Answer:

9

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