Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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 completely inelastic one and an elastic one,where the bullet bounces off the block. inelastic case elastic case a bullet inside no bullet inside A B A' B' What is the ratio of the flight time; i.e., tAB tA′B′ ?

Sagot :

onyebuchinnaji

The ratio of time of flight for inelastic collision to elastic collision 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

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/7694106

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.