Calculating the velocity of the center of mass of a system

[kunparekh18]

Hello everyone. First post here. I'll go the question straightaway.

Homework Statement

A gun of mass "M" placed on a smooth horizontal surface fires a bullet of mass "m" with a velocity "u" at an angle "θ" with the horizontal (ground). The velocity of the center of mass of the gun+bullet system after firing is (in terms of M,m,u and θ)?

Homework Equations

I know the velocity of the center of mass is given by the equation

v = mv / m

The Attempt at a Solution

I expanded the above equation to V = (Mv1 + mv2)/(M+m), where v2 and v1 are the velocities of masses M and m after firing.

However, I do not know what values to substitute for v1 and v2. I substituted v2 to be ucosθ as it is along the horizontal, and then used the conservation of linear momentum principle to get v1=-mucosθ/M

substituting these values if the original formula, I got a wrong answer. The final answer is supposed to be

musinθ/(M+m)

What mistake have I done here? Thanks in advance for explaining.

 

[TSny]

Hello, kunparekh18. Welcome to PF!

Did you consider separately the x (horizontal) and y (vertical) components of the velocity of the center of mass?

Since there are no external forces in the x direction, you should be able to deduce the final x-component of the velocity of the CM without any calculation.

 

[kunparekh18]

Hello, kunparekh18. Welcome to PF!

Did you consider separately the x (horizontal) and y (vertical) components of the velocity of the center of mass?

Since there are no external forces in the x direction, you should be able to deduce the final x-component of the velocity of the CM without any calculation.

Thanks!

I did consider the x and y components of the velocity of the center of mass separately.

As you said, there is no external force acting along the x axis, so I can apply conservation of linear momentum there.

0 + 0 = Mv1 + mucosθ

==> v1 = -(mucosθ)/M

So with that and my initial equation, I get the x component of the velocity of the center of mass to be 0.

Now there is one external force acting along the y axis, mg (weight of the bullet) (normal force by surface on M and weight Mg cancel out as there is no acceleration of M along y axis). How do I conserve momentum there?

By logic, I know that y component of velocity of M will be downwards in the -y direction, but since there is a surface there, there will be no displacement of the block in that direction. Hence, v1 along y-axis = 0. Using that and my first equation, I get the velocity of the center of mass to be

musinθ/M+m , which is perfectly correct. But how I do I derive that v1 along the y-axis is zero mathematically? I have got the answer, but I need a reasoning behind it (not for my homework, but for my own understanding. The question is multiple choice anyway :p )

 

[kunparekh18]

Bump?

 

[TSny]

I think your reasoning is fine.

The ground is a constraint that prevents the gun from moving in the negative y direction. So that's the reason you can set the final v_y for the gun equal to zero. I don't think there is a need to seek a mathematical reason.

 

[kunparekh18]

I think your reasoning is fine.

The ground is a constraint that prevents the gun from moving in the negative y direction. So that's the reason you can set the final v_y for the gun equal to zero. I don't think there is a need to seek a mathematical reason.

Thank you! I still have a doubt, what if the gun was shot downwards? Then the y component of the gun would be in the positive y direction? What would I do then? Thanks in advance.

 

[TSny]

I still have a doubt, what if the gun was shot downwards? Then the y component of the gun would be in the positive y direction? What would I do then? Thanks in advance.

What you would do in this case would depend on whether or not you can neglect the external impulse due to the force of gravity (weight) acting on the gun during the time of the explosion that propels the bullet.

If the gun is light enough compared to the explosive force of the firing, then you can ignore the external effect of gravity and use conservation of momentum to find the recoil velocity of the gun immediately after the firing. It would be like firing the gun in deep outer space.

If the gun is heavy, then it would not be a good approximation to assume conservation of momentum in the y-direction for the gun-bullet system. If the gun is heavy enough, then the gun might not even leave the ground when it is fired pointed downward. The effect of firing the gun would be to cause the normal force of the ground on the gun to decrease during the firing, but otherwise the gun would remain on the ground.

If the gun is light enough to leave the ground but still too heavy to ignore the gravitational force, then you would need to know more details about the explosive force of the firing in order to determine the speed at which the gun leaves the ground.