A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed .
Express your answer in terms of Vi,Vw, and Mb.
Vw+b = Vb x sqrt[Mb/(Mb+Mw)]
I've derived it from
0.5xMbx(Vb^2) = 0.5x(Mb+w)x(Vw+b)^2
To solve this problem, we can apply the principle of conservation of momentum.
Before the collision, the bullet is moving horizontally with speed Vi and has mass Mb, while the block is at rest and has mass Mw. After the collision, the bullet becomes completely embedded within the block, and they both move together with a final speed Vf.
According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Initial momentum = Final momentum
To calculate the initial momentum, we'll multiply the mass of the bullet (Mb) by its initial velocity (Vi):
Initial momentum = Mb * Vi
To calculate the final momentum, we'll multiply the combined mass of the bullet and the block (Mb + Mw) by their final velocity (Vf):
Final momentum = (Mb + Mw) * Vf
Since momentum is conserved, we can equate the two expressions:
Mb * Vi = (Mb + Mw) * Vf
Now, we can solve for Vf in terms of the given variables:
Vf = (Mb * Vi) / (Mb + Mw)
Therefore, the answer is Vf = (Mb * Vi) / (Mb + Mw)
Note: The mass of the wooden block, Mw, is not provided in the question, so we cannot express the final velocity solely in terms of this variable.