5) A 2 kg mass on the 2m ramp is allowed to slide on frictionless surface and co

Question

5) A 2 kg mass on the 2m ramp is allowed to slide on frictionless surface and collide elastically with 9kg mass. After collision, the two masses travel in together.  

a) What is the P.E. of 2kg mass on 2m ramp? ?(Hint: P.E. = mgh)

P.E. = mgh = 2kg *9.8m/s2 * 2m = 39.2 J

b) What is the velocity of the 2kg mass at the bottom of the ramp before collision?(Hint: K.E. = ½ m V2)

mgh = K.E. = ½ m V2 = 39.2 J V = = 6.26 m/s

I understand part A but I don't know how they got 6.26 m/s for the velocity. Could someone please help?

Explanation / Answer

See ,
Initial P.E of 2 Kg Mass = m*g*h
P.Ein = 2*9.8 * 2
P.E in = 39.2 J

Initial K.E = 0
Now, When Mass reaches bottom of the ramp -
Final P.E = 0
Final KE = 1/2 * m*V^2


Using Energy Conservation -
P.Ein + K.Ein = P.Efin + K.Efin
39.2 + 0 = 0 + 1/2 * m*v^2
39.2 = 1/2 * 2 * v^2
v = 6.26 m/s


So Just before Collision, 2 kg mass has a velocity of 6.26 m/s

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