Please help me...Physics problem?

Question

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring by 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

Answer 1

let M be the mass of block.let m be the mass of small block.

as F=kx, where F is force applied,k is spring constant, x is extension in spring., from initial data,we get

0.75=k(0.25/100)    [converting into SI units]

k=300N/m

as impact compresses it by .15cm,by energy conservation, we get

1/2[kx^2]=1/2[(m+M)v^2]

apply them and you will get

v=2.6 m/s(approx.)

Let initial speed of bullet be v' m/s

from momentum conservation, we can write

mv'=(m+M)v

we get,v'=325 m/s.

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Answer 2

a) Firstly we work out the stiffnesss of the spring..

F = kx

-0.75 = -k(0.25x10^-2)

=> k = 300 N/m

Now we use thstiffnessss to calculate the force applied by the block on the spring...

F = kx

F = -300(0.15)

=> F = -45N

Now we know F = ma {Newtons 2nd law of motion}

since the bullet is embedded we calculate a using total mass...

m = 0.008 + 0.992 = 1kg

F = ma

-45 = 1a

=> a = -45m/s^2

Use the equation of motion:

v^2 = u^2 + 2as  {v=velocity,u = initial velocity, a = acceleration, s = distance}

since the block stops for a moment when spring compresses...

v = 0

=> 0 = u^2 + 2(-45)(0.15)

u ~ 3.67m/s

b) we use the law of conservation of momentum

momentum before = momentum after

mv(bullet) = (m(bullet) + m(block)) v(block + bullet)

we know that v(block + bullet) = u in a above

v(block + bullet) ~ 3.67m/s

0.008v = 1(3.67)

v  ~ 459.28m/s

Hope this helps...

Cheers:)