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Anna11 [10]
2 years ago
7

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′ ?
Physics
1 answer:
pantera1 [17]2 years ago
4 0

The ratio of time of flight for inelastic collision to elastic collision is 1:2

The given parameters;

  • <em>mass of the bullet, = m₁</em>
  • <em>mass of the block, = m₂</em>
  • <em>initial velocity of the bullet, = u₁</em>
  • <em>initial velocity of the block, = u₂</em>

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

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

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

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A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m
Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

I_{disk} = \frac{1}{2} mr^2\\\\I_{disk}  = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk}  = 0.305 \ kgm^2

The moment of inertia of the rod about its center is calculated as follows;

I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2

The initial rotational kinetic energy of the disk and rod;

K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i=  \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \  \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J

The final rotational kinetic energy of the disk-rod system is calculated as follows;

K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J

The loss in rotational kinetic energy due to the collision is calculated as follows;

\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J  \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

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