A body moving horizontally on a smooth surface with a speed of $20$ $ms^{-1}$ splits into two parts and continue to move in the same direction. The masses of two parts are in the ratio of $1:2$. The smaller part moves with a speed of $30$ $ms^{-1}$. The fractional change is the kinetic energy is given by $\frac{1}{x}$. The value of x is__________
Problem Setup and Variables
Initial speed of the body, $v_i = 20 \, ms^{-1}$.
The body splits into two parts with masses in the ratio $1:2$. Let the initial mass be $M$. Then the masses of the two parts are $m_1 = \frac{M}{3}$ and $m_2 = \frac{2M}{3}$.
The smaller part ($m_1$) moves with speed $v_1 = 30 \, ms^{-1}$ in the same direction.
Let the speed of the larger part ($m_2$) be $v_2$. We need to find $v_2$ using conservation of momentum.
The fractional change in kinetic energy is given as $\frac{1}{x}$.
Momentum Conservation Analysis
The initial momentum of the body is $P_i = M \times v_i = M \times 20$.
The final momentum is the sum of the momenta of the two parts:
$P_f = m_1 v_1 + m_2 v_2 = (\frac{M}{3})(30) + (\frac{2M}{3})v_2 = 10M + \frac{2M}{3}v_2$.
Since the surface is smooth and there are no external forces in the horizontal direction, momentum is conserved ($P_i = P_f$):
$20M = 10M + \frac{2M}{3}v_2$
Subtract $10M$ from both sides:
$10M = \frac{2M}{3}v_2$
Solve for $v_2$:
$v_2 = \frac{10M \times 3}{2M} = 15 \, ms^{-1}$.
Kinetic Energy Calculation
Initial Kinetic Energy ($KE_i$):
$KE_i = \frac{1}{2} M v_i^2 = \frac{1}{2} M (20)^2 = \frac{1}{2} M (400) = 200M$.
Final Kinetic Energy ($KE_f$):
$KE_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$
$KE_f = \frac{1}{2} (\frac{M}{3}) (30)^2 + \frac{1}{2} (\frac{2M}{3}) (15)^2$
$KE_f = \frac{1}{2} (\frac{M}{3}) (900) + \frac{1}{2} (\frac{2M}{3}) (225)$
$KE_f = \frac{900M}{6} + \frac{450M}{6} = 150M + 75M = 225M$.
Fractional Change in Kinetic Energy
The change in kinetic energy is $\Delta KE = KE_f - KE_i = 225M - 200M = 25M$.
The fractional change in kinetic energy is:
Fractional Change $= \frac{\Delta KE}{KE_i} = \frac{25M}{200M} = \frac{1}{8}$.
We are given that the fractional change is $\frac{1}{x}$.
Therefore, $\frac{1}{x} = \frac{1}{8}$.
This implies $x = 8$.
Similar Questions
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Important Questions from Mechanics
- A soap bubble of surface tension $0.04 \text{ N/m}$ is blown to a diameter of $7 \text{ cm}$. If $(15000 - x) \text{ }\mu\text{J}$ of work is done in blowing it further to make its diameter $14 \text{ cm}$, then the value of $x$ is _______.
($\pi = 22/7$) - A uniform solid cylinder of length $L$ and radius $R$ has moment of inertia about its axis equal to $I_1$. A small co-centric cylinder of length $L/2$ and radius $R/3$ carved from this cylinder has moment of inertia about its axis equals to $I_2$. The ratio $I_1/I_2$ is _______.
- A fly wheel having mass $3\text{ kg}$ and radius $5\text{ m}$ is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to $3\text{ kg}$ mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by $3\text{ m}$ is _________ $\text{J}$. ($g = 10\text{ m/s}^{2}$)
Suppose there is a uniform circular disc of mass $M \text{ kg}$ and radius $r \text{ m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is ________.
- A ball of radius $r$ and density $\rho$ dropped through a viscous liquid of density $\sigma$ and viscosity $\eta$ attains its terminal velocity at time $t$, given by $t = A \rho^a r^b \eta^c \sigma^d$, where $A$ is a constant and $a$, $b$, $c$ and $d$ are integers. The value of $\frac{b+c}{a+d}$ is ________.
