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A block of mass M = 1.94 kg, at rest on a horizontal frictionless table, is...

A block of mass M = 1.94 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 110 N/m. A bullet of mass m = 4.7 g and velocity of magnitude 810 m/s strikes and is embedded in the block (Fig. See below). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

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Answer #1
Concepts and reason

The concept used to solve the question is the conservation of momentum and conservation of energy.

First, determine the final speed of the block or block-bullet combined system by using the concept of conservation of momentum. Later determine the amplitude of vibration by using the concept of conservation of energy, the kinetic energy of the body and total energy stored by the spring.

Fundamentals

Conservation of momentum:

The conservation of momentum is defined as the total momentum of the isolated system always remains constant or conserved.

The conservation of momentum for an isolated system is mathematically defined as,

(МУ), %3 (мУ),

Here, (MV)
and (MV)
are the initial and final momentum of the isolated system.

Conservation of energy:

The conservation of energy is defined as energy cannot be created nor destroyed only it can convert one form to another form of energy.

Write the expression for the kinetic energy of the object as:

К.Е 3D — mу?
2
2

Here, K.E
is the kinetic energy of the object, т
is mass of the object and V
is the speed of the object.

Write the expression for total energy stored by the spring as:

1
=
2

Here, is the total energy of the spring, k
is stiffness of the spring and is a maximum deflection of the object.

(a)

The conservation of momentum for an isolated system is defined as,

my, + MV, 3 (т+ М)у,

Here, т
is mass of the block, м
is mass of the block, is the initial speed of the bullet, is the initial speed of the block and is the final speed of the block.

Substitute 4.7g
forт
, 1.94kg
forм
, 810m/s
for and Om/s
for in above expression.

|(4.7g)(810m/s)+(1.94kg) (0m/s) ((4.7g)+(1.94 kg))V
(4.7g)(810m/s) (4.7 g) +(1.94 kg))
2
1kg
+(1.94 kg) |V2
1kg
(4.7g) 1000kg

Simplify the above expression for as

3.807 kg m/s
V2
1.9447 kg
=1.9576 m/s

(b)

According to the given condition after striking the bullet to the block, the kinetic energy of the bullet is fully converted into the oscillation energy of the spring.

The conservation of energy for bullet block and the spring system is mathematically defined as,

1
1
2.
kA2
mv
2
2

Here, A
is the amplitude of the spring.

The conservation of energy for bullet block and the spring system is mathematically defined as,

1
1
2.
kA2
mv
2
2

Substitute for ,810 m/s
for and 110 N/m
in the above expression a

(4.78)(810 m/s)(10N/m)4
1kg
4.781000kg810 m/s) =(110N/m)4
1.9035kg m2/s (55N/m)A

Simplify the above expression for A
,

2
1.9035kg m2/s
A =
(55N/m)
0.186m

Ans: Part a

The speed of the block immediately after the collision is1.9576 m/s
n
.

Part b

The amplitude of the resulting simple harmonic motion is0.186m
.

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