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# A bullet of mass 0.25 kg is fired with velocity 302 m/s into a block of wood of mass 37.5 kg .It gets embedded into it.The block m1 is resting on a block m2 and the horizontal surface on which it s placed is smooth.The coefficient of friction between m1 and m2 is 0.5.Find the displacement of m1 and m2 and the common velocity of m1 and m2.     Mass m2= 12.5 kg.Ans)  dispalcement = 0.013 and common velocity = 1.94 m/s and please give me the explanation of common velocity

8 years ago

Dear Rohit,

Let just after impact velocity of m1 is V1. Aceeleration of m1 and m2 are a1 and a2 respectively. Mass of bullet is m.

Apllying conservation of linear momentum, we find   V1=mv/(m1+m)= 2 m/s

a1=-Kg = -(0.5)(10)= - 5 m/s.s

a2= K(m+m1)g/m2= 15.1 m/s.s       where K is co-efficient of friction.

Let after t1 sec velocity of m1 and m2 equals, Relative velocity of m1 wrt m2  is represented by V12=0 at  t=t1 and V12=V1 at t=0

Hence V1/2=V1  + (a12)(t1)    where a12 is relative acceleration of m1 wrt m2

0=V1+(-20.1) t1

t1=0.0995 sec

Distance travelled by m1 during t1:

S1=V1 t1 -0.5a1(t1)^2=0.1742m

Distance travelled by m2 during t1

S2=0.5a2(t1)^2=0.0747 m

Relative displacement of m1  wrt to m2:

S12= S1 - S2

Commomn velocity:

V" = a2t1= 15.1*0.0995 = 1.5025 m/s

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