A block of mass m = 1.60 kg slides down a 30.0incline which is 3.60 m high. At t

Question

A block of mass m = 1.60 kg slides down a 30.0incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.10 kg which is at rest on a horizontal surface (Figure 1) . (Assume a smooth transition at the bottom of the incline.) The collision is elastic, and friction can be ignored.

PART A:Determine the speed of the block with mass m = 1.60 kg after the collision.

PART B:Determine the speed of the block with mass M = 7.10 kg after the collision.

PART C:Determine how far back up the incline the smaller mass will go.

Explanation / Answer

let the bottom of the incline be the referrence for zero potential energy.

initial potential energy=mass*g*height

this will get converted to kinetic energy when the block reaches the horizontal surface.

if speed at horizontal surface is v,

then 0.5*mass*v^2=mass*g*height

==>v=sqrt(2*g*height)=sqrt(2*9.8*3.6)=8.4 m/s

so just before the collison, the speed of block with mass 1.6 kg=8.4 m/s

if after collision, speed of smaller mass=v1 in opposite direction of its initial velocity

and speed of higher mass=v2 in direction of initial velocity of smaller mass

then using conservation of momentum principle :

1.6*8.4=-1.6*v1+7.1*v2

==>-1.6*v1+7.1*v2=13.44....(1)

==>v2=(13.44+1.6*v1)/7.1=1.893+0.22535*v1...(2)

using conservation of energy principle:

0.5*1.6*8.4^2=0.5*1.6*v1^2+0.5*7.1*v2^2

==>1.6*v1^2+7.1*v2^2=112.9

using equation 2:

1.6*v1^2+7.1*(1.893+0.22535*v1)^2=112.9

==>1.6*v1^2+7.1*(3.5834+0.050783*v1^2+0.85318*v1)=112.9

==>1.9606*v1^2+6.0576*v1-87.458=0

==>v1=-8.4 or v1=5.3104

ignoring the negative value,

v1=5.3104 m/s

then v2=1.893+0.22535*v1=3.0897 m/s

kinetic energy of the smaller mass=0.5*1.6*v1^2=0.5*1.6*5.3104^2=22.56 J

if maximum height reached is h,

then 1.6*9.8*h=22.56

==>h=1.4388 m

hence the smaller mass will climb up a height of 1.4388 m

distance climbed along the incline=height/sin(angle)

=1.4388/sin(30)=2.8776 m

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