Physics, 19.08.2019 21:30 alexisbreton

Arifle bullet with mass 8.00 g strikes and embeds itself in a block with a mass of .992 kg that rests on a frictionless, horizontal surface and it attached to a coil spring. the impact compresses the spring 15.0 cm. calibration of the spring shows that a force of 0.750 n is required to compress the spring 0.250 cm. find the magnitude of the block's velocity just after impact and the initial speed of the bullet.

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Answer from: lee9724

Ok, so this problem has to be solved in a couple of different steps. This involves conservation of energy and conservation of momentum.
Step 1, find the spring coefficient with the given information: F=kx.
0.75N=k*0.250cm (don't forget to convert units!) => 0.75 = k*0.0025m
k = 300 N/m.
Step 2, from the spring coefficient, you can see how much energy is stored in the spring from the impact. Spring Energy = 0.5*k*x^2 = 0.5*300*(0.15^2) = 3.375Nm
Step 3, using conservation of energy, the only thing the stored energy in the spring came from is the bullet before the impact. Nothing was moving before the bullet struck the block and the spring was relaxed. Kinetic Energy of the bullet = 0.5*m*v^2 = 0.5 * 0.008 * (v^2) = 3.375 (remember to use unit conversion to get everything into N, m, kg!) SO! v^2 = 843.75 and v = 29.05 m/s
Step 4, now that you have pre-impact velocity of the bullet, use conservation of momentum to find the post impact velocity of the bullet plus block. m_total*v_result = m_bullet*v_bullet + m_block*v_block In this case, the v_block =0 and v_bullet is the v we solved for in step 3.
Plugging in the numbers: m_total = 1, m_bullet = 0.008, v_bullet = 29.05
v_result = 0.008*29.05/1 = 0.2324 m/s

Answer from: tacoursey1987p6xy9s

The magnitude of the block's velocity just after impact is about 2.60 m/s.

The initial speed of the bullet is about 325 m/s

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of bullet = m₁ = 8.00 gram = 8 × 10⁻³ kg

mass of block = m₂ = 0.992 kg

initial speed of block = u₂ = 0 m/s

compression of spring = x = 15.0 cm = 0.15 m

spring constant = k = 0.750 ÷ (2.5 ×10⁻³) = 300 N/m

Asked:

final speed of block = v₂ = ?

initial speed of bullet = u₁ = ?

Solution:

Let:

final speed of the block just after impact = final speed of the bullet just after impact = v

Firstly , we will use Conservation of Energy as follows:

Ek = Ep

$\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2} k x^2$

(m_1 + m_2) v^2 = k x^2

$(0.008 + 0.992) v^2 = 300 \times 0.15^2$

v^2 = 6.75

$v = \frac{3}{2}\sqrt{3} \texttt{ m/s}$

$v \approx 2.60 \texttt{ m/s}$

The magnitude of the block's velocity just after impact is about 2.60 m/s

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Next , we will use Conservation of Momentum Law as follows:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

$0.008u_1 + 0.992(0) = 0.008(\frac{3}{2}\sqrt{3}) + 0.992(\frac{3}{2}\sqrt{3})$

$0.008u_1 = \frac{3}{2}\sqrt{3}$

$u_1 \approx 325 \texttt{ m/s}$

The initial speed of the bullet is about 325 m/s

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Learn moreImpacts of Gravity : Effect of Earth’s Gravity on Objects : The Acceleration Due To Gravity : Newton's Law of Motion: Example of Newton's Law:

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Arifle bullet with mass 8.00 g strikes and embeds itself in a block with a mass of .992 kg that rest

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