PHYSICS QUESTION: velocity and spring compression?

L

luvdmr

New member
A rifle bullet with mass 8.00 [g] strikes and embeds itself in a block with a mass of 0.992 [kg] that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 [cm]. Calibration of the spring shows that a force of 0.650 [N] is required to compress the spring 0.250 [cm].

Questions:
A) Find the magnitude of the block's velocity just after impact. [in m/s]

B) What was the initial speed of the bullet? [in m/s]
answer from let'slearntothink is correct.
thank you!
 
The energy of the spring at the end of the impact = 0.5*(0.650/0.0025)*(.15^2) = 2.925 J = 0.5 (M+m) v^2, where M is the mass of block, m is the mass of bullet and v is the velocity of block + bullet just after impact. So we have
v^2 = 2.925/[0.5*(0.992+0.008)] = 5.85 or v = 2.42 m/s^2

Let the initial speed of bullet be V. Momentum conservation gives
0.008 V = 1*v or V = 302.3 m/s. Therefore the velocity of the bullet before it struck was 302.3 m/s
 
first find the spring constant k = F/x = 0.65 / 0.25 x 10^-2 = 260 N/m

A) using conservation of energy law u have:

1/2(m + M) v^2 = 1/2 k x^2

2v^2 = 260 (0.15^2)

so v = 1.71 m/s
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B) using conserv. of momentum:

mu + 0 = (m + M)v

0.008 u = 2v

u = 2v / 0.008 = 2(1.71) / 0.008 = 427.5 m/s
 
The energy of the spring at the end of the impact = 0.5*(0.650/0.0025)*(.15^2) = 2.925 J = 0.5 (M+m) v^2, where M is the mass of block, m is the mass of bullet and v is the velocity of block + bullet just after impact. So we have
v^2 = 2.925/[0.5*(0.992+0.008)] = 5.85 or v = 2.42 m/s^2

Let the initial speed of bullet be V. Momentum conservation gives
0.008 V = 1*v or V = 302.3 m/s. Therefore the velocity of the bullet before it struck was 302.3 m/s
 
The velocity of the block+bullet after impact is found from conservation of momentum:

Vblt*Mblt = V*(Mblt + Mblk)

V = Vblt*Mblt/(Mblt + Mblk) = .008 Vblt

The energy bullet/block is 0.5*(Mblt + Mblk)*V²

KE = 0.5*(Mblt + Mblk)*[Vblt*Mblt/(Mblt + Mblk)]²

KE = 0.5*Vblt²Mblt²/(Mblt + Mblk)

The energy in the spring is PE = 0.5*k*∆x². The spring constant k = 0.650/0.250 = 2.6 N/cm = 2600 N/m, ∆x = 0.15 m, PE = 29.25 J

Equate the KE with the PE in the compressed spring:

29.25 = 0.5*Vblt²*Mblt²/(Mblt + Mblk)

58.5 = Vblt²*0.008²/1.00

Vblt = 959 m/s

The block's velocity is then 0.008*959 = 7.67 m/s
 
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