A 6.30g bullet traveling at 490m/s embeds itself in a 1.81kg wooden block at res
ID: 1902699 • Letter: A
Question
A 6.30g bullet traveling at 490m/s embeds itself in a 1.81kg wooden block at rest on a frictionless surface. The block is attached to a spring with k - 76.0N/m (see the figure). (Figure1) Find the period. Express your answer with the appropriate units. Find the amplitude of the subsequent simple harmonic motion. Express your answer with the appropriate units. Find the total energy of the bullet + block + spring system before the bullet enters the block. Express your answer with the appropriate units. Find the total energy of the bullet + block + spring system after the bullet enters the block. Express your answer with the appropriate units.Explanation / Answer
a)T=2sqrt(m/k)=0.97s
b)Conserving momentum,
mv=(M+m)V
V=1.70m/s
Conserving energy,
0.5(M+m)V2=0.5kx2
x=0.263m=26.3cm
c)Total energy before collision=0.5mv2=756.3J
d)Total energy after collison=0.5kx2=2.63J
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.