Question

A 12.7 g bullet is fired into a block of wood at 245 m/s. The block is attached to a spring constant of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block.

Answer 1

To solve this, we have to use two principles

1) Conservation of energy

2) Conservation of momentum

Let us denote

\(m_{1} m_{2}, M\) be the mass of bullet, mass of block of wood and mass of combined system (bullet + block of wood) respectively. \(u_{1}, u_{2}\) be the initial velocity of the bullet, block of wood and \(v\) be the velocity combined system. \(v_{i}, v_{f}\) be the initial and final velocity of the combined masses. \(x_{i}, x_{f}\) be the initial and final position of the spring.

The general idea is that the Kinetic energy of the combined system is transferred completed to the spring as potential energy when the velocity of the combined goes to zero.

Applying Conservation of energy

\(\frac{1}{2} M v_{i}^{2}+\frac{1}{2} K x_{i}^{2}=\frac{1}{2} M v_{f}^{2}+\frac{1}{2} K x_{f}^{2} \quad \rightarrow(1)\)

At maximum compression of \(35 \mathrm{~cm}_{2} v_{f}=0\) and at rest \(x_{i}=0 .\) The equation (1) becomes \(\frac{1}{2} M v_{i}^{2}=\frac{1}{2} K x_{f}^{2}\)

\(M=\frac{K x_{f}^{2}}{v_{i}^{2}}\)

\(\rightarrow(2)\)

Now use the principle of conservation of momentum. Momentum before collision = momentum after collision. \(m_{1} u_{1}+m_{2} u_{2}=M v\)

Since \(u_{2}=0\) \(M=\frac{m_{1} u_{1}}{v}\)

\(\rightarrow(3)\)

Equate equation \((2)\) and equation \((3)\), to calculate \(v_{i}\) Here \(v\) is denoted \(v_{i}\) i.e., as the initial velocity of the combined system.

\(\frac{K x_{f}^{2}}{v_{i}^{2}}=\frac{m_{1} u_{1}}{v_{i}}\)

\(v_{i}=\frac{K x_{f}^{2}}{m_{1} u_{1}}\)

\(=\frac{205 \times\left(35 \times 10^{-2}\right)^{2}}{12.7 \times 10^{-3} \times 245}\)

\(=8.07 \mathrm{~m} / \mathrm{s}\)

\(\rightarrow(4)\)

Substitute equation (4) in equation (3) \(M=\frac{m_{1} u_{1}}{v_{i}}\)

\(m_{2}=\frac{m_{1} u_{1}}{v_{i}}-m_{1}\)

\(=\frac{12.7 \times 10^{-3} \times 245}{8.07}-12.7 \times 10^{-3}\)

\(=373 \mathrm{~g}\)

Therefore, the mass of the block of wood is \(373 \mathrm{~g}\)