You are at the controls of a particle accelerator, sending a beam of 1.50 X 10 m/s protons (mass ?m?) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.20 X 10 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic. (a) Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass ?m.? (b) What is the speed of the unknown nucleus immediately after such a collision?

Solution 50E Step 1: a) According to law of conservation of momentum, we can say that, Momentum before collision = momentum after collision Momentum before collision = momentum of proton before collision + momentum of the nuclei before collision Momentum of proton before collision = m v , where, v - spep p proton p 7 Provided, speed of proton, v = 1.50 × 1p m/s Momentum of unknown nuclei before collision = m v x x Provided, the initial velocity of unknown nuclei is negligible. So, v = 0 m/s x then , Momentum of unknown nuclei before collision = m v = 0 x x Therefore, momentum before collision = Momentum of proton = m v = 1.50 × 10 m/s × 7 p p m p Momentum after collision = momentum of proton after collision + momentum of the nuclei after collision Momentum of proton after collision = m v’ , where, v - peedpf proton p 7 Provided, speed of proton after collision, v’ = 1.20 × 10 m/p Momentum of unknown nuclei after collision = m v’ x x 7 Therefore, momentum after collision = m v’ +m v’ = (1.20 × 10 m/s) × m + m p p x x p x x 7 7 That is, 1.50 × 10 m/s × m = (1.20 × 10 m/s) × m +p v’ p x x Rearranging, m v’ = 3× 10 m/s m 6 x x p 6 v’ x(3× 10 m/s) m /m p x Step 2: Since the collision is elastic, the energy is conserved The KE before collision = The KE after collision KE before collision = ½ m v = ½ m (1.50 × 10 m/s) 2 7 2 p p p 14 2 2 KE before collision = (2.25m /2) × 10 m /s p KE after collision = ½ m v’ + ½ m v’ 2 2 p p x x Substituting the value for v’ from step 1 x KE after collision = ½ m (1.20 × 10 m/s) + ½ m (3 × 10 ) m /m 7 2 6 2 2 2 p p p x 14 2 2 7 2 12 2 2 Therefore, (2.25m /2) × 10 m /s p½ m (1.20 × 10 m/s) + ½ m ×9× 10 ×(m /mp) p p x Cutting the term ½ on both sides 14 2 2 14 2 2 14 2 2 3 2 2.25m × 10p /s = 1.44 m × 10 m /s + 0.09 × 10 mp/s (m /m ) p x Cutting the term 10 14 and unit m /s on both sides 3 2 2.25m = 1.4pm + 0.09 × (m p ) p x Cutting m on bothpides 2 2 2.25 = 1.44 + 0.09 (m /m ) p x Rearranging, 2.25 - 1.44 = 0.09 (m /m ) p x2 0.09 (m /m ) = 0.81 2 p x Or, (m /m ) = 0.81/0.09 = 9 p x Taking square root on both sides (m /p ) =x That is, m = m /x p rd Mass of the unknown nuclei is 1/3 of proton.