Collisions+in+two+dimensions

Hi Tish i'm having quite a lot of trouble with the fourth part of the collisions booklet Momentum. i've tried the activity 4A and was able to work out the first question but questions 2 and 3 i'm finding very difficult and i can't quite work out how to use the vector diagrames to work out the solutions i've checked the answers but i still don't understand how to get their answers. Is their a method that will work for each question? if so could you tell me.

Question 2 A body of mass 4 kg travels north and strikes a stationary body of 6 kg. After the collision, the 4kg body moves off North East at a velocity of 6kg. What is the velocity of the 6 kg body after the collision?

solution I always **draw diagrams**, which is a bit hard in an email  One object is stationary, v = 0 and hence, total momentum before collision is just 4kg x 10m/s = 40 kgm/s North. This is also the total momentum after the collision. But now the 4 kg mass is travelling at 6 m/s north east, so **its** momentum after the collision is 24 kgm/s North east. To find the momentum of the 6 kg mass you have to do a vector subtraction. That is subtract 24 kgm/s North east from 40 kgm/s North.  40 kgm/s North **minus** 24 kgm/s North east   Draw the vector diagram to scale and complete the triangle. The third side points sort of North west (actually 36o west of north) and is 27 kgm/s long. So the velocity of the 6 kg mass after collision = 27 ÷ 6 = 4.5 m/s. I’m a little under the answer given because of my scale drawing, say 5 m/s to the nearest whole number.  These kind of collisions may not be in the standards, but vector subtraction certainly is – mostly for changes of velocity
 * Vector subtraction ** means you have to point the arrow for the value subtracted in the **opposite direction**, in this case south west.

 Question 3 a neutron, m = 1.7 x10-27, travelling at 6 x 106 m/s, collides with another particle which is stationary. The stationary particle has an unknown mass. After colliding the neutron changes direction by 45 degrees and slows down. The unknown particle travels at right angle to the neutron at a velocity of 1.5 x 106 m/s. What is the mass of the unknown particle? Question 3 works much the same way. <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">But it has more complicated values in it. But because it has the two particles moving off at right angles to each other after the collision, you can use **Pythagoras’** theorm. <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;"> <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">Total momentum before collision = 1.7x10-27 kg x 6x106 m/s = 1.02x10-20 kgm/s <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;"> <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">This is also the total momentum **after the collision.** <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">After the collision, the momentum of the neutron = momentum of the unknown particle and the triangle is a right angle triangle <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">So <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;"> (momentum of the neutron)2 + (momentum of the unknown particle)2 = (1.02x10-20 kgm/s)2 <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;"> <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">Both particles move off at 45o, so this is an **isosceles** right angle triangle, **their momentums** after collision are **equal.** <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">Solving this momentum of the unknown particle = 7.2x1021 <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">Now to find the mass of the unknown particle = momentum ÷ its velocity = 7.2x1021 ÷ 1.5x106 m/s <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;"> = 4.8x10-27 kg
 * <span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;"> 2 x **<span style="color: #1f497d; font-family: "Calibri","sans-serif"; font-size: 11pt;">(momentum of the unknown particle)2 = (1.02x10-20 kgm/s)2