Question

A bullet of mass 1.7×10−3 kg embeds itself in a wooden block with mass 0.980 kg , which then compresses a spring (k = 160 N/m ) by a distance 6.0×10−2 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.53. What is the initial speed of the bullet?

Answer 1

Step 1:

Find Initial speed of bullet + block system after collision

Using energy conservation:

KEi + PEi + Wf = KEf + PEf

KEf = 0, since final speed of system is zero

PEi = 0, since initially no compression in spring

KEi = (1/2)*M*V^2

M = combined mass of bullet + spring = m1 + m2 = 1.7*10^-3 + 0.980 = 0.9817 kg

V = initial speed of system just after collision = ?

PEf = Potential energy due to compression in spring = (1/2)k*d^2

k = spring constant = 160 N/m

d = compression = 6.0*10^-2 m = 0.06 m

Wf = Work-done by friction force = Ff*d

Wf = -uk*N*d = -uk*M*g*d

So,

(1/2)*M*V^2 + 0 - uk*M*g*d = 0 + (1/2)*k*d^2

V^2 = k*d^2/M + 2*uk*g*d

Using given values:

V = sqrt [160*0.06^2/0.9817 + 2*0.53*9.81*0.06]

V = 1.10 m/sec

Step 2:

Using Momentum conservation find initial speed of bullet:

Pi = Pf

m1*u1 + m2*u2 = M*V

u2 = initial speed of block = 0 m/sec

u1 = initial speed of bullet = ?

So,

u1 = (M*V - m2*u2)/(m1)

u1 = (0.9817*1.10 - 0.980*0)/(1.7*10^-3)

u1 = initial speed of bullet = 635.2 m/sec

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