A(n) 8.5-kg object is sliding across the ice at 2.34 m/s in the positive x direc

Question

A(n) 8.5-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 11 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is afrontx, the x component fo the average acceleration of the other chunk during the explosion is arearx

Explanation / Answer

lets assume that the other chunk moves in negative x direction.

let the final velocity (after 0.16 s time period) of first chunk is v1 and second chunk is v2.

v1 is along +ve x direction and v2 is along -ve x direction

then using conservation of momentum principle:

8.5*2.34=8.5*0.5*v1-8.5*0.5*v2

==> v1-v2=4.68 ...(1)

using conservation of energy:

initial kinetic energy+addition of energy=final kinetic energy

==>0.5*8.5*2.34^2+11=0.5*0.5*8.5*v1^2+0.5*8.5*0.5*v2^2

==>v1^2+v2^2=16.1276

using v1=v2+4.68,

we get

v2^2+4.68^2+9.36*v2+v2^2=16.1276

==>2*v2^2+9.36*v2+5.7748=0

solving for v2, we get v2=-0.7312 m/s

or v2=-3.9488 m/s

taking v2=-0.7312 m/s, v1=3.9488 m/s

taking v2=-3.9488 m/s, v1=0.7312 m/s

so in either case, both the blocks are moving in same direction (+ve x direction) with one moving with speed 3.9488 m/s and other moving with 0.7312 m/s

average acceleration=change in velocity/time taken

so afrontx=(3.9488-2.34)/0.16=10.055 m/s^2

arearx=(0.7312-2.34)/0.16=-10.055 m/s^2

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