A(n) 7.8-kg object is sliding across the ice at2.34 m/s in the positive x direct

Question

A(n) 7.8-kg object is sliding across the ice at2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 12 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 xdirection. 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.

What are the x components of the average accelerations of the two chunks during the explosion?

Enter your answers numerically separated by a comma.

Explanation / Answer

The velocity of object moving forward=2.34+xm/s^2

The velocity of object moving backward=2.34-xm/s^2

The converging energy is

1/2mv^2+12J=1/2(m/2)(v1)^2+1/2(m/2)(v2)^2

1/2mv^2+12J=1/2(m/2)(2.34+x)^2+1/2(m/2)(2.34-x)^2

1/2m(2.34)^2+12j=1/2(m/2)((2.34)^2+x^2+2(2.34)(x))+1/2(m/2)((2.34)^2+x^2-2(2.34)(x))

1/2m(2.34)^2+12J=1/2m(2.34)^2+1/2mx^2

1/2mx^2=12J

X=root((12*2)/m)

X= root((12*2)/7.8)=1.753m/s

The velocity of the first body=2.34+xm/s^2

=2.34+1.753m/s

=4.093m/s

Acceleration is v=u+at

                     4.093=(2.34)+a(0.16)

                     a=4.093-2.34/0.16

a front x =10.95m/s^2

The velocity of the second body =2.34-xm/s^2

=2.34-1.753m/s^2

=0.587m/s

Acceleration=v-u/t=0.587-2.34/0.16

a rear x =-10.95m/s^2

(a front, a rear)=(10.95, -10.95)

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