Question

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′ ?

Answer 1

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;

$m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)$

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

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

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

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\m_1u_1 + 0 = m_1v_1 + m_2v_2\\\\m_1 u_1 = m_1v_1 + m_2v_2$

Apply one-directional velocity

$u_1 + (-v_1) = u_2 + v_2\\\\u_1 -v_1 = 0 + v_2\\\\v_1 = v_2 -u_1$

Substitute the value of $v_1$ into the above equation;

$m_1u_1 = m_1(v_2 - u_1) + m_2 v_2\\\\m_1u_1 = m_1v_2 -m_1u_1 + m_2v_2\\\\2m_1u_1 = m_1v_2 + m_2v_2\\\\2m_1u_1= v_2(m_1 + m_2)\\\\v_2 = \frac{2m_1u_1}{m_1+ m_2} \ --(3)$

where;

$v_2$ 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;

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

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

$\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2$

Learn more about elastic and inelastic collision here: brainly.com/question/7694106