inelastic collision

Question

A massless spring (with force constant k = 183 N/m) connects a wall and a block of wood. The system is initially at rest, with the spring unstretched. The block has mass M = 51.7 g and is able to move without friction on a table. A gun is positioned to fire a bullet of mass m = 6.5 g into the block along the spring axis. After the gun is fired, the bullet gets embedded in the block, and the spring is compressed a maximum distance d = 0.81 m.

 Find an equation for the speed of the bullet v, just before it hits the block, in terms of the variables given in the problem statement. 

Where did the M+m come from and why was d/m factored out?

✔ Correct!

Answer 1

Because energy is conserved, we know that the kinetic energy of the block+bullet has to equal the energy from the compressed spring. This gives the equation

(1/2)(m+M)V² = (1/2)kd²

where V is the speed of both the block and bullet... and it's unknown. We can find this V is by using conservation of momentum: the momentum before the bullet hits the block is equal to the momentum after the bullet hits the block (remember, momentum is p=mv). This gives the equation

m*v = (m+M)V

-> V = m*v/(m+M)

Now we can plug V into the first equation, and solve for v

(1/2)(m+M)m²*v²/(m+M)² = (1/2)kd²

-> m²*v²/(m+M) = kd²

-> v² = kd²(m+M)/m²

-> v = √(kd²(m+M)/m²) = √[(d²/m²)(m+M)k] = (d/m)√(k(m+M))