Respuesta :
Answer:
A. 1.172 metres
B. 6.82 Ns
C. 4.796 m/s
Explanation:
The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.
The 1.40 kg block is at rest so velocity is zero and has no momentum.
The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s
Momentum = 310*0.022
Momentum = 6.82 Ns.
If the bullet gets embedded they will both have common velocity v
6.82 = (0.022+1.40)v
6.82 = 1.422v
V = 6.82/1.422
V = 4.796 m/s
How high the block will rise after the bullet is embedded is given by
H = (U²Sin²tita)/2g
Where tita is 90°
H = (4.796² * sin²(90))/(2*9.81)
H =( 23.001616*1)/19.62
H = 1.172 metres
(a) The maximum height reached by the bullet block system is 1.18 m.
(b) The initial momentum of the system is 6.82 kgm/s.
(c) The speed of the block just after the bullet becomes embedded in it is 4.8 m/s.
The given parameters;
- mass of the block, m₁ = 1.4 kg
- mass of the bullet, m₂ = 22 g = 0.022 kg
- initial velocity of the bullet, u₂ = 310 m/s
The final velocity of the bullet block system is calculated as;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\1.4(0) + 0.022(310) = v(1.4 + 0.022)\\\\6.82 = v(1.422)\\\\v = \frac{6.82}{1.422} \\\\v = 4.8 \ m/s[/tex]
The maximum height reached by the bullet block system is calculated as;
[tex]P.E = K.E\\\\(m_1+m_2) gh = \frac{1}{2} (m_1 + m_2)v^2\\\\gh = \frac{v^2}{2} \\\\h = \frac{v^2}{2g} \\\\h = \frac{4.8^2 }{2\times 9.8} \\\\h = 1.18 \ m[/tex]
The initial momentum of the system is calculated as;
[tex]P_i = m_1u_1 + m_2u_2\\\\P_i = 1.4(0) + 0.022(310)\\\\P_i = 6.82 \ kgm/s[/tex]
The speed of the block just after the bullet becomes embedded in it = 4.8 m/s.
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