Question

Q.1 Please explain how

$m _{A}u _{A} + m_{B}u _{B} = m _{A}v_{A} + m_{B}v_{B}$
Came from

$m _{A} \frac{(v _A - u _A)}{t} = m _B \frac{(v _B - u _B)}{t}$
Q.2 A bullet of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2kg. What is recoil velocity of the pistol ?
(Use the above formula to take out the answer)

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Answer 1

1. first answer on attachment
2. According to law of conservation of linear momentum,
4. MV = mv.
6. M is mass of pistol = 2 kg.
8. m is mass of bullet = 2 g = 2 (10 ^-3) kg.
10. v = velocity of bullet = 150 m/s.
12. V is recoil velocity of pistol . V= (m/M) v = [ 2(10^-3)/(2)] (150) = 0.15 m/s

Answer 2

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QUESTION 1 :-

$m_A \frac{(v_A - u_A)}{t } = - m_B \frac{(v_B - u_B)}{t}$

t will be cut because there are two t as denominators.

$m_Av_A - m_Au_A = - m_Bv_B + m_Bu_B$

$m_Av_A + m_Bv_B = m_Av_A + m_Bu_B$

QUESTION 2 :-

A bullet is of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2kg. What is the recoil velocity of the pistol ?

Using the above formula

$m_A = 20g = \frac{20}{100} = 0.02kg$

$m_B = 2kg$

$u_A = 0$

$u_B = 0$

$v_A = 150m/s$

$v_B = ?$

$m_Av_A + m_Bv_B = m_Au_A + m_Bu_B$

Putting values :-

$0.02 \times 150 + 2 \times v_B = 0.02 \times 0 \times 0$

$3 + 2v_B = 0$

$2v_B = 0 - 3$

$2v_B = - 3$

$v_B = \frac{ - 3}{ \: \: 2}$

$v_B = - 1.5m/s$

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