The force exerted by object X on Y is 4F to the right and the force exerted by object Y on X is F to the left.
The given parameters;
The force exerted by the object Y is calculated as follows;
[tex]F = \frac{mv}{t} \\\\[/tex]
where;
t is the time of impact
[tex]F_Y = \frac{MV_0}{t}[/tex]
The force exerted by the object X is calculated as follows;
[tex]F = \frac{mv}{t} \\\\[/tex]
[tex]F_X = \frac{2M \times 2V_0}{t} \\\\F_X = \frac{4MV_0}{t} \\\\F_X = 4(\frac{MV_0}{t} )[/tex]
[tex]F_X = 4(F_Y)[/tex]
Thus, we can conclude that the force exerted by object X on Y is 4F to the right and the force exerted by object Y on X is F to the left.
Learn more here:https://brainly.com/question/13710974
The force exerted by object X on Y is 4F to the right. And the force exerted by object Y on X is F to the left. Hence, option (a) is correct.
Given data:
The mass of object X is, 2M.
The initial velocity of X is, [tex]2v_{0}[/tex].
The mass of object Y is, M.
The initial velocity of Y is, [tex]v_{0}[/tex].
Applying the concept of impulse-momentum to obtain average force by object X as,
[tex]F=\dfrac{(2M)(2v_{0})}{t}\\F=\dfrac{4Mv_{0}}{t}[/tex]............................................................(1)
Here, t is the reaction time.
Similarly, average force by the object Y is,
[tex]F'=\dfrac{Mv_{0}}{t}[/tex]...................................................................(2)
Taking ratio of equation (1) and (2) as,
[tex]\dfrac{F}{F'}=\dfrac{\dfrac{4Mv_{0}}{t}}{\dfrac{Mv_{0}}{t}}\\\dfrac{F}{F'}= 4\\F = 4F'[/tex]
Thus, we can conclude that the force exerted by object X on Y is 4F to the right. And the force exerted by object Y on X is F to the left. Hence, option (a) is correct.
Learn more about the impulse-momentum theorem here:
https://brainly.com/question/14121529?referrer=searchResults
