Answer to Question #198769 in Mechanics | Relativity for karen

Question

A stone whose mass is 0.1 kg rests on a horizontal frictionless surface. A bullet of mass 0.0025 kg, traveling horizontally at 400 m/s, strikes the stone and rebounds horizontally at right angle to its original direction with a speed of 300 m/s. Compute the magnitude and direction of the velocity of the stone after it is struck

Answer 1

Gives

$m_1=0.0025kg$

$m_2=0.1kg$

$u_1=400m/sec$

$u_2=0m/sec$

$v_1=300m/sec$

X-axis components

$$m_1u_1+m_2u_2=m_1u_1cos\theta+m_2u_2cos\phi$$

$$0.0025\times400+0.1\times0=0.0025\times300cos90°+0.1\times v_2cos\phi$$

$1=0+0.1v_2cos\phi$

$v_2cos\phi=10\rightarrow(1)$

Y -axis components

$0+0=m_1v_1sin\theta-m_2v_2sin\phi$

$$0=0.0025\times 300\times sin90°-0.1\times v_2sin\phi$$

$0.75-0.1v_2sin\phi$

$v_2sin\phi=7.5\rightarrow(2)$

Eqution (2)/equation (1)

$\frac{v_2sin\phi}{v_2cos\phi}=\frac{7.5}{10}$

$tan\phi=0.75$

$\phi=36.86°$

Put value in eqution (1)

$10=v_2cos(36.86°)$

$v_2=12.49m/sec$