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Math  /  Algebra

QuestionProblem 6: ( 10%10 \% of Assignment Value) A block with mass M=59.2 gM=59.2 \mathrm{~g} rests on a frictionless wooden table. On its right side it is attached to the wall with a massless linear spring with a force constant k=183 N/mk=183 \mathrm{~N} / \mathrm{m}. Initially the block is at rest, and the spring is in its equilibrium position. A bullet with mass m=8.43 gm=8.43 \mathrm{~g} and an unknown horizontal velocity of magnitude vv strikes and becomes embedded in the block causing the spring to compress by a maximum distance d=0.802 md=0.802 \mathrm{~m}. - Part'(a) V\boldsymbol{V}
Select the option that correctly expresses the initial speed of the bullet. v=dmk(M+m)v=\frac{d}{m} \sqrt{k(M+m)}
Correct!
Part (b) What is the speed, in meters per second, of the bullet just before it enters the block? v=v= □\square m/s\mathrm{m} / \mathrm{s} Grade Summary Deductions Potential 0%0 \% 100\% Submissions Attempt(s) Remaining: 5 0%0 \% Deduction per Attempt detailed view Submit Hint Feedback I give up! 5 Submission(s) Remaining Hints: 0‾\underline{0} for a 0%0 \% deduction. Hints remaining: 0‾\underline{0} Feedback: 0%‾\underline{0 \%} deduction per feedback. All content © 2024 Expert TA, LLC

Studdy Solution

STEP 1

1. The block and bullet system is isolated and frictionless.
2. The spring is initially at equilibrium and compresses by d=0.802 m d = 0.802 \, \text{m} .
3. The bullet becomes embedded in the block, forming a single mass after the collision.
4. The conservation of momentum and energy principles apply.

STEP 2

1. Apply conservation of momentum to find the velocity of the block-bullet system immediately after the collision.
2. Use conservation of energy to relate the kinetic energy of the block-bullet system to the potential energy stored in the spring.
3. Solve for the initial speed of the bullet.

STEP 3

Apply conservation of momentum. Before the collision, the momentum is due to the bullet only. After the collision, the momentum is shared by the block and bullet together.
Initial momentum: m⋅v m \cdot v
Final momentum: (M+m)⋅V (M + m) \cdot V
Set them equal:
m⋅v=(M+m)⋅V m \cdot v = (M + m) \cdot V

STEP 4

Apply conservation of energy. The kinetic energy of the block-bullet system is converted into the potential energy of the spring.
Kinetic energy after collision:
12(M+m)V2 \frac{1}{2} (M + m) V^2
Potential energy in spring at maximum compression:
12kd2 \frac{1}{2} k d^2
Set them equal:
12(M+m)V2=12kd2 \frac{1}{2} (M + m) V^2 = \frac{1}{2} k d^2

STEP 5

Solve for V V from the energy equation:
(M+m)V2=kd2 (M + m) V^2 = k d^2
V2=kd2M+m V^2 = \frac{k d^2}{M + m}
V=kd2M+m V = \sqrt{\frac{k d^2}{M + m}}

STEP 6

Substitute V V back into the momentum equation to solve for v v :
m⋅v=(M+m)⋅kd2M+m m \cdot v = (M + m) \cdot \sqrt{\frac{k d^2}{M + m}}
v=(M+m)m⋅kd2M+m v = \frac{(M + m)}{m} \cdot \sqrt{\frac{k d^2}{M + m}}
v=dm⋅k(M+m) v = \frac{d}{m} \cdot \sqrt{k(M + m)}
The speed of the bullet just before it enters the block is:
v=0.8020.00843⋅183×(0.0592+0.00843) v = \frac{0.802}{0.00843} \cdot \sqrt{183 \times (0.0592 + 0.00843)}
v≈220.5 m/s v \approx 220.5 \, \text{m/s}

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