A rifle bullet with mass 9.0 grams strikes and embeds itself into a block with m

Question

A rifle bullet with mass 9.0 grams strikes and embeds itself into a block with mass 992 grams that rests on a frictionless, horizontal surface. The block is attached to a spring. The impact causes the spring to compress 15.0 cm before stopping. A previous calibration of the spring found that it takes a force of 0.750 Newtons to compress the spring 0.25 cm Find the magnitude of the block's velocity immediately after impact What was the initial speed of the bullet? What was the impulse of the bullet on the block if it took 0.05 seconds for the bullet to embed itself into the block?

Explanation / Answer

Calculating spring constant of spring ,
We know,
F = k*x
0.75 = k * 0.25/100
k = 300 N/m

Using Energy conservation,
Kinetic Energy of Block = Spring Potential Energy
1/2* (m+M)*v^2 = 1/2*k*x^2
(9 +992)/1000 * v^2 = 300 * 0.15^2
v = 2.60 m/s
Block velocity immediately after impact, v = 2.60 m/s

(b)
Let the initial spped of bullet be vb

Using Momentum Conservation,
Initial Momentum = Final Momentum
m*vb = (m+M) * v
9*vb = (9 + 992) * 2.6
vb = 289 m/s

(c)
t = 0.05 s
Impulse, I = F*t
F = k*x = 300 * 0.15 = 45 N
I = 45 * 0.05 Ns
I = 2.25 Ns

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