Question

Two objects $A$ and $B$ are thrown upwards simultaneously with the same speed. The mass of $A$ is greater than the mass of $B$. Suppose the air exerts a constant and equal force of resistance on the two bodies. Which object will go high?

Answer 1

Hint: When an object is thrown vertically upward it is under the action of three forces.

(a) The force with which it is thrown

(b) The gravitational force acting on it due to its mass.

(c) Air pressure

For solving such kinds of problems we always draw the free body diagram of the object.

${F_a}$: The force with which the object is thrown vertically upward

${F_g}$: Gravitational Force

${F_r}$: Force due to air pressure.

Complete step by step solution:

Here we have got two objects A and B. We are given that the Mass of object A (${M_A}$) is greater than the mass of object B (${M_B}$). Now we know that Force (F) is

$F = M \times a$

Here (M) is the mass of the object and (a) is acceleration.

In this case (a) is deceleration due to gravity because the objects are going upward and the gravitational pull is in the downward direction.Now, it is given that both of air exerts constant and equal force on both the bodies. So we can say that ${F_r}$ is equal for both the bodies

$\therefore {\left( {{F_r}} \right)_A} = {\left( {{F_r}} \right)_B}$

\[\Rightarrow {M_A} \times {a_A} = {M_B} \times {a_B} \]

\[\because {M_A} > {M_B} \]

\[ \therefore {a_A} < {a_B} \]

From equation of motion

${v^2} = {u^2} + 2as$

Where, S: distance covered by the object, u: initial velocity, v: final velocity and a: acceleration.

Here it is already told that the initial speed (u) of both the objects are the same.One the object reaches the topmost point its final velocity becomes zero for a moment and then gradually starts increasing as it comes down so we are considering the topmost point of both the objects where the final velocities of both the object will be zero

\[\therefore 0 = {u^2} + 2as \]

\[ \Rightarrow {u^2} = - 2as \]

\[ \Rightarrow s = - \dfrac{{{u^2}}}{{2a}} \]

Here the values of (a) will be negative as it is deceleration or negative acceleration which will cancel out the minus sign of the above equation.

$\Rightarrow s = \dfrac{{{u^2}}}{{2a}}$

Now as the value of initial velocity (u) is the same for both the objects so we can say that

Distance covered (s) is inversely proportional to the acceleration.

$\Rightarrow s \propto \dfrac{1}{a}$

Now, as the acceleration of object A is less than object B. From the above condition, distance traveled or height achieved by object A will be more than object B.

${S_A} > {S_B}$

Therefore, object A with higher mass will go higher than object B with lesser mass.

Note: 

Sign of the force is decided by its directions.

- Negative acceleration or retardation is always negative.

- The value of acceleration due to gravity is fixed and is $9.81\dfrac{m}{s}$.

- A free-body diagram is a diagram where all the forces acting on the body are shown simultaneously.