Question

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless horizontal surface. Sam, with a mass of 80.0 kg, is given a push and slides eastward. Abigail, with a mass of 50.0 kg, is sent sliding northward. They collide and, after the collision, Sam is moving at 6.00 m/s in a direction 37.0° north of east, while Abigail is moving at 9.00 m/s in a direction 23.0° south of east. Find the speeds of Sam and Abigail just before the collision.

Answer 1

1. Answer: Velocity of Sam =
   `9.97ms^(-1)` and velocity of Abigail =
   `2.26 ms^(-1)` Explanation: Consider Vs as sam velocity and Va as bigail velocity. Now consider the north as positive x and the east as positive y for our reference. Before collision, their velocity vectors can be represented as
   `Vs_(1)` = Vs i
   `Va_(1)` =Va j Now after collision their velocity vectors are given as
   `Vs_(2)` = 6 cos 37 i + 6 sin 37 j = 4.79 i + 3.61 j
   `Va_(2)` = 9 cos 23 i + 9 sin 23 j = 8.28 i - 3.52 j As we know in the absence of external force the momentum before and after collision will remain the same therefore P1 = P2 ∴ P = mv Momentum = mass x velocity
   `m_(1) v_(1) =m_(2)v_(2)` As sliding is on frictionless surface therefore before and after the collision, momentum remains conserved in the absence of external force. we can write it as
   `m_(s) v_(s1) + m_(a)v_(a1)  = m_(s)v_(s2)  +m_(a) v_(a2)` Now putting values we get 80 x Vs + 50 x 0 = 80 x 4.79 + 50 x 8.28 (First Consider horizontal vector components) Velocity of sam before collision= Vs = 9.97
   `ms^(-1)` Now consider vertical vector components 80 x 0 + 50 x Va = 80 x 3.61 + 50 x (-3.52) (Before Collision direction opposite so put negative sign) velocity of abigail before collision = Va = 2.26
   `ms^(-1)`