
Force F acts on a body of mass 1 kg moving with an initial velocity \[{v_0}\] for 1 sec. Then
A. Distance covered by the body is \[{v_0} + \dfrac{F}{2}\]
B. Final Velocity of the body is \[{v_0} + F\]
C. Momentum of the body is increased by F
D. All of the above
Answer
163.2k+ views
Hint: Use equations of motion for a uniformly accelerated body to find the distance covered and final velocity of the body. Momentum of a body is the quantity used to describe the state of the motion of an object having non-zero mass and is equal to the product of mass and velocity.
Formula used :
\[s = ut + \dfrac{1}{2}a{t^2}\] and v= u + at
Here, s = Distance covered, u = Initial velocity, a = Acceleration, v = Final velocity and t = time
Complete step by step solution:
Given here, is a body of mass 1 kg moving with an initial velocity \[{v_0}\] for 1 sec.
Force = F, Mass = 1kg
Initial velocity = \[{v_0}\], time t = 1 sec.
Distance covered by the body can be calculated by using equation,
\[s = ut + \dfrac{1}{2}a{t^2}\]
Where, u = initial velocity and a = acceleration,
\[a = \dfrac{F}{m} = F\]............(As, m = 1 kg)
Therefore,
\[s = {v_0}(1) + \dfrac{1}{2}F{\left( 1 \right)^2} \\
\Rightarrow S = {v_0} + \dfrac{F}{2}\]
Final velocity can be calculated by equation,
v= u + at
Where, v = final velocity and u = Initial velocity
Therefore,
\[v = {v_0} + F(1) \\
\Rightarrow v = {v_0} + F\]
Momentum P of the body is equal to the product of mass m and velocity v of the body.
As the initial velocity of the body is therefore initial momentum will be,
\[{P_i} = m{v_0} = {v_0}\] and,
Final momentum will be,
\[{P_f} = m\left( {{v_0} + F} \right)\]
Therefore, change in momentum will be \[\Delta P = {P_f} - {P_i}\]
\[\Delta P = m\left( {{v_0} + F} \right) - {v_0} = F\]
Momentum of the body will increase by F.
Therefore, option D is the correct answer.
Note: Initial and Final velocity of an object is thought of as the first and last velocity of the object but they can be better described as the velocity at which the observation is done for the first and last time.
Formula used :
\[s = ut + \dfrac{1}{2}a{t^2}\] and v= u + at
Here, s = Distance covered, u = Initial velocity, a = Acceleration, v = Final velocity and t = time
Complete step by step solution:
Given here, is a body of mass 1 kg moving with an initial velocity \[{v_0}\] for 1 sec.
Force = F, Mass = 1kg
Initial velocity = \[{v_0}\], time t = 1 sec.
Distance covered by the body can be calculated by using equation,
\[s = ut + \dfrac{1}{2}a{t^2}\]
Where, u = initial velocity and a = acceleration,
\[a = \dfrac{F}{m} = F\]............(As, m = 1 kg)
Therefore,
\[s = {v_0}(1) + \dfrac{1}{2}F{\left( 1 \right)^2} \\
\Rightarrow S = {v_0} + \dfrac{F}{2}\]
Final velocity can be calculated by equation,
v= u + at
Where, v = final velocity and u = Initial velocity
Therefore,
\[v = {v_0} + F(1) \\
\Rightarrow v = {v_0} + F\]
Momentum P of the body is equal to the product of mass m and velocity v of the body.
As the initial velocity of the body is therefore initial momentum will be,
\[{P_i} = m{v_0} = {v_0}\] and,
Final momentum will be,
\[{P_f} = m\left( {{v_0} + F} \right)\]
Therefore, change in momentum will be \[\Delta P = {P_f} - {P_i}\]
\[\Delta P = m\left( {{v_0} + F} \right) - {v_0} = F\]
Momentum of the body will increase by F.
Therefore, option D is the correct answer.
Note: Initial and Final velocity of an object is thought of as the first and last velocity of the object but they can be better described as the velocity at which the observation is done for the first and last time.
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