Question

An ice block is melting at constant rate $$\left|{\displaystyle \frac{dm}{dt}}\right|=\mu$$. Its initial mass is $$m_0$$ and it is moving with a velocity $$v_0$$ on a frictionless horizontal surface. The distance travelled by it till it melts completely is:
A. $${\displaystyle \frac{2m_0{v}_0}{\mu }}$$
B. $${\displaystyle \frac{m_0{v}_0}{2\mu }}$$
C. $${\displaystyle \frac{m_0{v}_0}{\mu }}$$
D. Can’t be said

Answer 1

Hint: Ice block is moving with increasing velocity due to decrease in the mass. There is no external force acting on the ice block. Use Newton’s second law of motion and law of conservation of momentum to determine the acceleration of the ice block. Use kinematic relation to determine the distance travelled by the ice block.

Formula used:
The momentum of the body of mass m is,
$$p=mv$$, where, v is the velocity.
According to Newton’s second law of motion,
$$F={\displaystyle \frac{dp}{dt}}$$
 $$v^2=v_0^2+2as$$
Here, v is the final velocity, $$v_0$$ is the initial velocity, a is the acceleration and s is the distance travelled.

Complete step by step answer:
We have given the rate of melting of the ice block, $${\displaystyle \frac{dm}{dt}}=-\mu$$. The negative sign represents the mass of the ice is decreasing.
To calculate the distance travelled by the ice block, we need the acceleration of the block. We can see as the mass of the ice block decreases, the velocity increases as there is no external force acting on it. Therefore, according to law of conservation of momentum, we can write,
$$m_0{v}_0=mv$$
Here, m is the mass and v is the velocity of the block as any instant of time.
According to Newton’s second law of motion, we can write,
$$F={\displaystyle \frac{dp}{dt}}=0$$ …………… (Since the net force on the block is zero)
Here, p is the momentum of the block.
Since the momentum is the product of mass and velocity, we can express the above equation as follows,
$${\displaystyle \frac{dp}{dt}}=m{\displaystyle \frac{dv}{dt}}+v{\displaystyle \frac{dm}{dt}}=0$$
$$\Rightarrow m{\displaystyle \frac{dv}{dt}}=-v{\displaystyle \frac{dm}{dt}}$$
$$\Rightarrow ma=-v\left(-\mu \right)$$ ………… (Since$$a={\displaystyle \frac{dv}{dt}}$$)
$$\Rightarrow a={\displaystyle \frac{v\mu }{m}}$$
We know that when no external force acts on the body, the acceleration of the body remains constant. Therefore, the initial acceleration of the ice block is equal to the final acceleration.
$$\Rightarrow a={\displaystyle \frac{v_0\mu }{m_0}}$$
We can use kinematic equation to calculate the distance travelled by the ice block as follows,
$$v^2=v_0^2+2as$$
Here, v is the final velocity, $$v_0$$ is the initial velocity, a is the acceleration and s is the distance travelled.
At the final point when the ice block completely melts, its final velocity becomes zero. Therefore, we can write,
$$v_0^2=\left|2as\right|$$
$$\Rightarrow s={\displaystyle \frac{v_0^2}{2a}}$$
We substitute $${\displaystyle \frac{v_0\mu }{m_0}}$$ for a in the above equation.
$$s={\displaystyle \frac{v_0^2}{2\left({\displaystyle \frac{v_0\mu }{m_0}}\right)}}$$
$$\Rightarrow s={\displaystyle \frac{m_0{v}_0}{2\mu }}$$

So, the correct answer is “Option B”.

Note:
For a body moving with increasing velocity, the acceleration of the body can never be negative. In the above kinematic equation, we have taken the magnitude of the distance and neglected the negative sign. To solve such types of questions, students need to understand where the final and initial velocity can be zero.