A block of mass 0.3 kg is attached to a spring of spring constant 30 N/m on a fr

Question

A block of mass 0.3 kg is attached to a
spring of spring constant 30 N/m on a fric-
tionless track. The block moves in simple har-
monic motion with amplitude 0.12 m. While
passing through the equilibrium point from
left to right, the block is struck by a bullet,
which stops inside the block.
The velocity of the bullet immediately be-
fore it strikes the block is 52 m/s and the mass
of the bullet is 2.06 g.
Find the speed of the block immediately
before the collision.
Answer in units of m/s
If the simple harmonic motion after the colli-
sion is described by x = B sin(t + ), what
is the new amplitude B?
Answer in units of m
The collision occurred at the equilibrium po-
sition.
How long will it take for the block to reach
maximum amplitude after the collision?
Answer in units of s

Explanation / Answer

a) first we need to figure out the velocity of the block. The velocity of the block at the equilibrium point will be the product of the angular frequency and the amplitude

=(k/m)=(30/.3)

=10

v=A=10(.12)

v=12 m/s

Now we use conservation of momentum

.3(12)+(.00206)(52)=(.3+.00206)v

v=12.273 m/s

b)Now we have to look at energy where all the energy is KE and PE and set them equal

PE=KE

(1/2)kB2=(1/2)mv2

B2=(m/k)v2=(.30206/30)(12.2732)

B=1.5166 m

c)The time from equilibrium to max amplitude it 1/4 the period.

T=2/=2/10)

T=..628

t=T/4

t=..1571 seconds

Hope that helps

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