SIR\MADAM PLEASE EXPLAIN ME ABOUT THIS PROBLEM

Asked by  | 4th Nov, 2009, 11:09: PM

Expert Answer:

Let m1 be tha mass of bullet and m2 that of block, then,

Total mass = 0.012 + 0.1 =0.112 kg

Now, μ = 0.650 and g = 9.8 m s-2

Force of friction, F = μ R = 0.650 x 0.112 x 9.8 N

So, acceleration, a = F / m = 6.37 m s-2

As the block comes to rest,  this is the retardation.

If u1 is the initial velocity of the bullet, then

K.E. of bullet = (1/2) m1u12

This energy is tansferred to the block which moves with an initial velocity say V , then

(1/2) m1u12 = ( 1/2) M V2

or, V2 = (0.012 / 0.112 ) u12

           =  0.107 u12

As the block along with the bullet comes to rest after covering 7.5 m , its final velocity becomes zero ,

So, - 2 x 6.37 x 7.5 = V2 = 0.107 u12

or, u = 29.8 m /s

Answered by  | 5th Nov, 2009, 03:48: PM

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