Calculate the velocity of the players just after the tackle

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To calculate the velocity of two players after a perfectly inelastic collision, use the conservation of momentum principle. The total momentum before the collision is equal to the total momentum after, leading to the equation 90(-5) + 95(3) = (90 + 95)v'. After determining the common final velocity, the decrease in total kinetic energy can be calculated using the kinetic energy equations before and after the collision. The initial kinetic energy is the sum of both players' kinetic energies before the tackle, while the final kinetic energy is based on their combined mass and the calculated final velocity. This approach effectively demonstrates the principles of momentum and energy conservation in collisions.
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Need Help Please!

A 90kg fullback moving south with a speed of 5m/s has a perfectly inelastic collision with a 95kg opponent running north at 3m/s.

a. Calculate the velocity of the players just after the tackle

b. calculate the decrease in total kinetic energy as a result of the collision.
 
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Perfectly inelastic means they stick together so they have a common final velocity. Use the principle of conservation of momentum and the kinetic energy equation for the 2nd one. Show me how you think you would do this question bearing in mind what I've said.
 
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Well, if they have a total inelastic collision then after the collision they move together, as one body, with the same velocity, right?
So all you've got to do is use the momentum conservation law to compare the total momentum before the collision and after it and then find the velocity.

About b, after you've calculate the new velocity then \Delta E_{k} = E_{k\ total\ after\ collision} - E_{k\ total\ before\ collision}
 
if u call north the positive direction then

<br /> m_1v_1 + m_2v_2 = (m_1 + m_2)v&#039; <br />

simple conservation of momentum...

<br /> 90(-5) + 95(3) = 185v&#039;<br />

now u can get the velocity just after the collision...
and if u use the kinetic energy equation...

<br /> E_{before} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2<br />

...

<br /> E_{after} = \frac{1}{2}(m_1+m_2)v&#039;^2

;)
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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