Question

(25%) Problem 3: A bullet is fired horizontally into an initially stationary block of wood suspended by a string and remains embedded in the block. The bullet’s mass is m = 0.0085 kg, while that of the block is M = 0.99 kg. After the collision the block/bullet system swings and reaches a maximum height of h = 0.95 m above its initial height. Neglect air resistance.

Answer 1

Answer:

Thus, the initial velocity of the bullet is 507.5 m/s.

Explanation:

mass of bullet, m = 0.0085 kg

mass of block, M = 0.99 kg

Height raised, h = 0.95 m

Let the initial velocity of bullet is u and the final velocity of block and bullet system is v.

Use conservation of energy

Potential energy of bullet bock system = kinetic energy of bullet block system

$(M+m)\times g\times h = \frac{1}{2}\times \left ( M+m \right )v^{2}$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 0.95}=4.32 m/s$

Now use conservation of linear momentum

mu = (M+m) v

0.0085 x u = (0.99 + 0.0085) x 4.32

0.0085 u = 4.314

u = 507.5 m/s

Thus, the initial velocity of the bullet is 507.5 m/s.  

Answer 2

The initial velocity of the bullet is 507.4 m/s

Your question is not complete, it seems to be missing the following information;

find the initial velocity of the bullet

The given parameters;

• mass of the bullet, m₁ = 0.0085 kg

• mass of block, m₂ = 0.99 kg

• maximum height attained by the system, h = 0.95 m

The final velocity of the system is calculated as;

$\frac{1}{2} (m_1 + m_2) v_{max}^2 = (m_1 + m_2) gh_{max}\\\\v_{max}^2 = 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}} \\\\v_{max} = \sqrt{2\times 9.8 \times 0.95} \\\\v_{max} = 4.32 \ m/s$

The initial velocity of the bullet is calculated by applying the principle of conservation of linear momentum;

$m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\0.0085u_1 + 0 = 4.32(0.0085 + 0.99)\\\\0.0085u_1 = 4.313\\\\u_1 = \frac{4.313}{0.0085} \\\\u_1 = 507.4 \ m/s$

Thus, the initial velocity of the bullet is 507.4 m/s

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