Answer:
The initial velocity of the bullet is 1066.63 m/s
Explanation:
The problem involves two different events: A collision between two objects and the stretching of a spring
Since we know the final condition of the system, we start from the end and back to the initial condition
The system of the bullet+block with mass= 4Kg+8gr = 4.008 Kg stretches the spring a distance of x=0.087 m
The work done when stretching a spring of constant k is
[tex]W=\frac{1}{2}kx^2=\frac{1}{2}(2400N/m)(0.087m)^2=9.0828 J[/tex]
This work is equivalent to the change of potential energy of the spring, which is the result of the transformation of the kinetic energy of the total mass of the system
[tex]KE=\frac{1}{2}m'v'^2[/tex]
Solving for v' (the velocity of the system after the collision took place)
[tex]v'=\sqrt{\frac{2KE}{m}}=2.129 m/s[/tex]
This is the resulting velocity after the bullet lodged into the block. That event complies with the law of conservation of linear momentum
[tex]m_{bullet}v_{bullet}+m_{block}v_{block}=m_{total}v'[/tex]
Since the block is initially at rest ([tex]v_{block}=0[/tex])
[tex]m_{bullet}v_{bullet}=m_{total}v'[/tex]
Solving for [tex]v_{bullet}[/tex]:
[tex]v_{bullet}=\frac{m_{total}v'}{m_{bullet}}=\frac{4.008Kg 2.129 m/s}{0.008Kg}=1066.63 m/s[/tex]
The initial velocity of the bullet is 1066.63 m/s
The initial velocity of the bullet is mathematically given as
v=1065.5m/s
Question Parameters:
An 8.0-g bullet is shot into a 4.0-kg block
The block moves into an ideal massless spring and compresses it by 8.7 cm.
The spring constant of the spring is 2400 Nm
Generally the equation for the conservation of Momentum is mathematically given as
mV=(m+M)v1
where, conservation of Momentum
1/2(m+m)v^2=1/2(m+m)m^2/(m+m)^2*v^2
=1/2kx^2
Hence,
[tex]v=\sqrt{k(m+m)x/m}\\\\v=sqrt{2400*4*\frac{0.087}{0.008}}[/tex]
v=1065.5m/s
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