Question

In an elementary-particle experiment, a particle of mass \(m\) is fired, with momentum \(m c,\) at a target particle of mass \(2 \sqrt{2} m .\) The two particles form a single new particle (in a completely inelastic collision). Find the following: a) the speed of the projectile before the collision b) the mass of the new particle c) the speed of the new particle after the collision

Short answer

Answer: The speed of the projectile before the collision is \(v=c\), and the speed of the new particle after the collision is \(V=\frac{c}{1+2\sqrt{2}}\).

Step-by-step solution

Determine Initial Momentum and Energy

We'll start by finding the total initial momentum and kinetic energy of the two particles before the collision. The initial momentum is given by \(m c\) for the projectile (particle of mass \(m\)) and \(0\) for the target particle (since it's stationary). The initial kinetic energy is \((1/2)m v^2\) for the projectile, where \(v\) is the initial speed we want to find. Total initial momentum: \(m c\) Total initial kinetic energy: \((1/2)m v^2\) (we need to find \(v\))

Step 2a: Find the Speed of the Projectile Before the Collision

To find the speed of the projectile before the collision, we need to rewrite the momentum given, \(m c,\) in terms of the speed \(v\): \(m c = m v \Rightarrow c = v\) The speed of the projectile before the collision is \(v = c\).

Step 2b: Find the Mass of the New Particle

In a completely inelastic collision, the masses of the two particles add up to form the mass of the new particle. So, the new particle's mass will be: \(M = m + 2\sqrt{2}m = (1 + 2\sqrt{2})m\)

Step 2c: Determine the Speed of the New Particle After the Collision

Since momentum is conserved in the collision, the total initial momentum will equal the total final momentum. The final momentum is given by \(MV\), where \(M\) is the mass of the new particle from Step 2b and \(V\) is the speed of the new particle after the collision. Total final momentum: \(MV\) Using conservation of momentum: \(m c = MV\) We can solve for the speed of the new particle \(V\): \(V = \frac{m c}{M} = \frac{m c}{(1 + 2\sqrt{2})m} = \frac{c}{1 + 2\sqrt{2}}\) The speed of the new particle after the collision is \(V = \frac{c}{1 + 2\sqrt{2}}\).