A bullet fired into a hard target losses half of it's K.E after 3cm.how much distance will it cover before coming to rest.assuming it constantly experience opposing force?

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Answer 1 · 2018-06-25T16:45:40

The further distance travelled before coming to rest is #=3cm#

Let the speed on entering the target is #=u#

The mass of the bullet is #=m#

The kinetic energy on entering the target is

#KE_1=1/2m u^2#

Let the speed after entering the target by #s_1=0.03m# is #=v#

Then,

The kinetic energy on entering the target is

#KE_2=1/2mv^2#

Therefore,

#(1/2mv^2)/(1/2m u^2)=0.5#

Therefore,

#v^2=0.5u^2#

Apply the equation of motion

#v^2=u^2+2as_1#

#a=(v^2-u^2)/(2s_1)=(0.5u^2-u^2)/(2*0.03)#

#=-0.5u^2/0.06#

Let the final velocity be #v_1=0#

And the total distance travelled be #=s_2#

Then,

#v_1^2=u^2+2as_2#

#0=u^2+2*(-0.5u^2/0.06)*s_2#

#2*(0.5u^2/0.06)*s_2=u^2#

#s_2=0.06/(2*0.5)=0.06m#

The further distance travelled before coming to rest is

#=s_2-s_1=0.06-0.03=0.03m#