a) From what distance d was the 0.9 kg block released? The 0.9 and 1.8 kg blocks

Question

a) From what distance d was the 0.9 kg block released?

The 0.9 and 1.8 kg blocks stick together when they are hit.

b) what is the speed of the combined blocks just after the collision?

The combined blocks slide up the second incline and stop 0.16 meters above the floor level.

c) How much work was done by the friction force as they slide up the incline?

A 0.9 kg block is released from rest on a frictionless 35 degree incline, starting at a distance d up the incline. A 1.8 kg block is at rest on the frictionless floor. Beyond the 1.8 kg block is another incline which is not frictionless. The speed of the 0.9 kg block just before it hits the 1.8 kg block is 6.2 m/s. a) From what distance d was the 0.9 kg block released? The 0.9 and 1.8 kg blocks stick together when they are hit. b) what is the speed of the combined blocks just after the collision? The combined blocks slide up the second incline and stop 0.16 meters above the floor level. c) How much work was done by the friction force as they slide up the incline?

Explanation / Answer

A. As the speed of block = 6.2 m/s

Since there is no friction, use conservation of energy: Loss in potential energy = Gain in K.E

Loss in potential energy = m*g*h = m*g* (d *Sin(35) ) = 0.9 *9.81 * (d *Sin(35) )

Gain in K.E = 1/2 mV2 = 1/2 *0.9 * 6.22

So, 0.9 *9.81 * (d *Sin(35) ) = 1/2 *0.9 * 6.22

d = 0.35 m <--------------ans

B. Blocks stick together

Applying Consevation of momentum as no external force is acting.

initial momentum = Final

0.9 * 6.2 = (0.9+1.8) Vcombined

Vcombined = 2.07 m/s <-----------------ans

C.stop 0.16 meters above the floor level

So, Gain in potential energy = Mgh =( 0.9 +1.8) * 9.81 * 0.16 =4.24 joule

Initial K.E =1/2 * M * V2combined = 5.79 joule

So applying Consevation of energy

Initial K.E =Gain in potential energy +Work done by friction

Work done by friction = 5.79 -4.24 = 1.55 joule <----------ans

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