Question

A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is ?k.
Part A
Find an expression for the bullet's speed vbullet.
Part B
What is the speed of a 9.0 g bullet that, when fired into a 12 kg stationary wood block, causes the block to slide 5.4 cm across a wood table? Assume that ?k=0.20.

Answer 1

A. The bullet's speed is [ ( M + m ) / m ] √ ( 2 μ g d )

B. The initial speed of the 9.0 g bullet is about 610 m/s

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Further explanation

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

where:

I = impulse on the object ( kg m/s )

∑F = net force acting on object ( kg m /s² = Newton )

t = elapsed time ( s )

Let us now tackle the problem!

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Given:

mass of bullet = m = 9.0 g = 9.0 × 10⁻³ kg

mass of block = M = 12 kg

sliding distance = d = 5.4 cm = 5.4 × 10⁻² m

coefficient of kinetic friction = k = 0.20

Asked:

initial bullet's speed = u₁ = ?

Solution:

Part A :

Firstly, we will use Conservation of Energy formula to find the speed of the block:

$W = \Delta Ek$

$fd = \frac{1}{2}(M+m)v^2$

$\mu N d = \frac{1}{2}(M+m)v^2$

$\mu (M + m)g d = \frac{1}{2}(M+m)v^2$

$\mu g d = \frac{1}{2} v^2$

$\boxed {v = \sqrt{2 \mu g d}}$ → Equation A

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Next, we will use Conservation of Momentum formula to find the initial speed of the bullet:

$\texttt{Total Momentum Before Collision = Total Momentum After Collision}$

$m u_1 + M u_2 = ( m + M ) v$

$m u_1 + M (0) = ( m + M ) v$

$m u_1 = ( m + M ) v$

$m u_1 = ( m + M ) \sqrt { 2\mu g d}$ ← Equation A

$\boxed {u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}}$

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Part B :

$u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}$

$u_1 = \frac { 9.0 \times 10^{-3} + 12 }{ 9.0 \times 10^{-3} } \sqrt { 2 \times 0.20 \times 9.80 \times 5.4 \times 10^{-2}}$

$\boxed{u_1 \approx 610 \texttt{ m/s}}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer 2

(a) The expression for the speed of the bullet is $u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}$

(b) The speed of the bullet at the given parameters is 613.78 m/s.

The given parameters;

• mass of the bullet, = m
• mass of the block, = M
• distance traveled, = d
• coefficient of friction, = k

Apply the principle of conservation of linear momentum to determine the initial speed of the bullet;

$m_1 u_1 + m_2 u_2 = V(m_1 + m_2)\\\\m_1 u_1 + 0 = V(m_1 + m_2)\\\\u_1 = \frac{V(m_1 + m_2)}{m_1}$

Apply the principle of work-energy theorem to determine the speed of the bullet-block system;

$K.E - P.E = W_f\\\\\frac{1}{2} MV^2 - 0 = \mu_k (Mg)d\\\\V^2 = 2\mu_k gd\\\\V = \sqrt{2\mu_k gd$

The expression for the speed of the bullet is written as;

$u_1 = \frac{V(m_1 + m_2)}{m_1} \\\\u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}$

The speed of the bullet at the given parameters is calculated as follows;

$u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1} \\\\u_1 = \frac{(0.009 + 12) \sqrt{2\times 0.2 \times 9.8 \times 0.054} }{0.009} \\\\u_1 = 613.78 \ m/s$

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