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A bullet of mass M1, is fired towards a block
of mass m2 initially at rest at the edge of a
frictionless table of height h as in the figure.
The initial speed of the bullet is vi. Consider
two cases. a соmрlеtеlу inсlаѕtiс one and an
elastic one where the bullet bounces off the block. What is the flight ratio time?

Sagot :

onyebuchinnaji

The ratio of time of flight for inelastic collision to elastic collision [tex](t_A :t_B)[/tex] is 1:2

The given parameters;

  • mass of the bullet, = m₁
  • mass of the block, = m₂
  • initial velocity of the bullet, = u₁
  • initial velocity of the block, = u₂ = 0

Considering inelastic collision, the final velocity of the system is calculated as;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)\\\\[/tex]

The time of motion of the system form top of the table is calculated as;

[tex]v = u + gt\\\\v = 0 + gt\\\\v = gt\\\\t= \frac{v}{g} \\\\t_A = \frac{m_1u_1}{g(m_1 + m_2)} \ \ ---(2)[/tex]

Considering elastic collision, the final velocity of the system is calculated as;

[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\m_1u_1 + 0 = m_1v_1 + m_2v_2\\\\m_1 u_1 = m_1v_1 + m_2v_2[/tex]

Apply one-directional velocity

[tex]u_1 + (-v_1) = u_2 + v_2\\\\u_1 -v_1 = 0 + v_2\\\\v_1 = v_2 -u_1[/tex]

Substitute the value of [tex]v_1[/tex] into the above equation;

[tex]m_1u_1 = m_1(v_2 - u_1) + m_2 v_2\\\\m_1u_1 = m_1v_2 -m_1u_1 + m_2v_2\\\\2m_1u_1 = m_1v_2 + m_2v_2\\\\2m_1u_1= v_2(m_1 + m_2)\\\\v_2 = \frac{2m_1u_1}{m_1+ m_2} \ --(3)[/tex]

where;

[tex]v_2[/tex] is the final velocity of the block after collision

Since the bullet bounces off, we assume that only the block fell to the ground from the table.

The time of motion of the block is calculated as follows;

[tex]v_2 = v_0 + gt\\\\v_2 = 0 + gt\\\\t = \frac{v_2}{g} \\\\t_B = \frac{v_2}{g} \\\\ t_B = \frac{2m_1u_1}{g(m_1 + m_2)} \ \ ---(4)[/tex]

The ratio of time of flight for inelastic collision to elastic collision is calculated as follows;

[tex]\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2[/tex]

Learn more about elastic and inelastic collision here: https://brainly.com/question/12497950

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