A 20.0 g bullet strikes a 0.608 kg block attached to a fixed horizontal spring whose spring constant is 6.66 ×?
A 20.0 g bullet strikes a 0.608 kg block attached to a fixed horizontal spring whose spring constant is 6.66 × 103 N/m and sets it into oscillation with an amplitude of 21.0 cm. What was the initial speed of the bullet if the two objects move together after impact?
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The speed of the block + bullet immediately after impact is
= mbu*vbu/(mbu+mbl) = 0.02vbu/(0.628 kg) = 0.031847vbu.
The kinetic energy of the (block + bullet) is
Ekin = 1/2 mv^2 = 1/2*(0.628)*(0.031847vbu)^2
This energy is equal to the spring energy when compressed 0.21 m:
Espring = 1/2*k*s^2 = 1/2*6.66*10^3*0.21^2
1/2*(0.628)*(0.031847vbu)^2 = 1/2*6.66*10^3*0.21^2
vbu = 679.06 m/s <--- velocity of bullet
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this could be an exceedingly cool question i like physics.!! First momentum is conserved: m1v1 = m2v2;m1= bullet v1= bullet m2= m1+block v2=new velocity This new velocity is haas a kinetic skill .5m2v2^2 and all of that kinetic skill would be switched over to skill skill interior the spring U=.5kx^2 while it reaches the amplitude. m1 = .038 kg m2 = .938 kg V1 =? amp = .18 m ok = 6* 10^3 So: v2 = m1v1/m2 --> KE = .5m2(m1v1/m2)^2 = .5 m1^2 v1^2 /m2 = .5Kx^2 ; locate v1 v1^2 = ok x^2 m2 / (m1^2 ) sq. root the two factors to get: 126,3 hundred m/s the belief of the subject is robust undecided abotu the mathematics section artwork it out and notice.