1. 5.00-kg mass and a 9.00-kg mass are being held at rest against a compressed s

Question

1. 5.00-kg mass and a 9.00-kg mass are being held at rest against a compressed spring on a frictionless surface. When the masses are released, the 5.00-kg mass moves to the east with a speed of 2.00 m/s What is the velocity of the 9.00-kg mass after the masses are released? 2. You (50.0-kg mass) skate on ice at 5.0 m/s to greet your friend (40.0-kg mass), who is standing still, with open arms. As you collide, while holding each other, with what speed do you both move off together? How much kinetic energy is lost in this collision? 3. Two identical freight cars roll toward one another, on a level track with negligible friction. One car is moving at 2.3 m/s and the other car at 2.7 m/s. After the collision, both cars are coupled and roll together with what speed? How much kinetic energy is lost in this collision?

Explanation / Answer

1)

Using law of conservation of linear momentum

Linear momentum of the system before motion = Linear momentum of the system after motion

before motion the bodies are at rest

then

pi = pf

(m1*0)+(m2*0) = (m1*v1)+(m2*v2)

0 = (5*2)+(9*v2)

10 = -9*v2

v2 = -10/9 = -1.11 m/sec

i.e 9 kg body move with velocity 1.11 m/sec to west

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2) using law of conservation of linear momentum

(50*5)+(40*0) = (50+40)*v

250 = 90*v

v = 250/90

v = 2.78 m/sec

initial kinetic energy is Ki = 0.5*m1*vi^2 = 0.5*50*5^2 = 625 J

Final kinetic energy is Kf = 0.5*(m1+m2)*v^2 = 0.5*(50+40)*2.78^2 =347.778 J

change in kinetic energy is Kf-Ki = 347.77-625 = -277.23 J

ask question 3) as a seperate question

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