Linear Motion and Linear Momentum

[Delta2]

According to the Impulse-Momentum Theorem the change in the momentum of a system of bodies is equal to the total impulse of the external forces acting on the system.
So:
Change in momentum of the system bullet+block $\Delta P=(M+m)u-mv$
Impulse of external forces acting on the system: The external forces are the weight of the bullet $mg$ and the weight of the block $Mg$. Their impulses , by definition, are equal to $-mg\Delta t$ and $-Mg\Delta t$ (we take as positive the up direction, that is the direction of the initial velocity of the bullet $v$).
So by impulse-momentum theorem we have $$(M+m)u-mv=-mg\Delta t-Mg\Delta t$$ and the final equation follows with just a tiny bit of algebra.

 

[LordJulius]

According to the Impulse-Momentum Theorem the change in the momentum of a system of bodies is equal to the total impulse of the external forces acting on the system.
So:
Change in momentum of the system bullet+block $\Delta P=(M+m)u-mv$
Impulse of external forces acting on the system: The external forces are the weight of the bullet $mg$ and the weight of the block $Mg$. Their impulses , by definition, are equal to $-mg\Delta t$ and $-Mg\Delta t$ (we take as positive the up direction, that is the direction of the initial velocity of the bullet $v$).
So by impulse-momentum theorem we have $$(M+m)u-mv=-mg\Delta t-Mg\Delta t$$ and the final equation follows with just a tiny bit of algebra.

okayy i understand it now thank you so much