A block (m=.75kg) is released from the top of a frictionless inclineed plane tha

Question

A block (m=.75kg) is released from the top of a frictionless inclineed plane that has a height of 1.5m and makes an angle of 50 degrees with respect to the horizontal. The inclined plane is attached to a horizontal surface. A second black (M=2.75 kg) is at rest on the horizontal surface at a distance 2.5m from the bottom of the inclined plane in the path of the first block. After the collision, the first block bounces back and comes to rest at a distance of 1m from the point of collision. The coefficient of kinetic friction between the blocks and the horizontal surface is .12.

A) How far from the point of collision will the second block come to rest

B)How much energy was lost in the collision?

Explanation / Answer

m*g*h = 0.5*m*v^2 ;

v = sqrt (2*g*h)

h= 1.5

v = 5.422 m/s ;

Work done by friction = mew*m*g*s = 0.12*0.75*9.8*2.5 = 2.205 J

Final velocity of Block A before collision = ???

0.5*m*v^2 = 0.5*m*5.422^2 - 2.205

v = 4.8495 m/s ;

Now Let after collision velocity of Block A be va and of block B be vb ;

0 = va^2 - 2*a*s

s = 1 m

a = mew*g = 0.12*9.8 = 1.176 m/s^2

va = 1.5336 m/s (in opposite direction )

now conserving momentum :

0.75*4.8495 = 0.75*-1.5336 + 2.75*vb

vb = 1.7408 m/s ;

A)

0 = vb^2 - 2*a*s

a = 1.176 m/s^2

s = vb^2/(2a) = 1.2885 m (ans to A part)

B) Energy lost = final Energy - Initial energy

= 0.5*0.75*1.5336^2 + 0.5*2.75*1.7408^2 - 0.5*0.75*4.8495^2

= 3.77 J (ans to Part B)

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