Question

Part A A 14 kg projectile is launched at 380 m/s at a 55 angle to the horizontal. At the peak of its trajectory it collides with a second projectile moving horizontally, in the opposite direction, at 140 m/s The two stick together and land 9.9 km horizontally downrange from the first projectile's launch point Find the mass of the second projectile. Express your answer using two significant figures. m-9.3 kg Submit Previous Answers Request Answer X Incorrect: Try Again; 4 attempts remaining

Answer 1

Vertical { Horizontal components are V_1x = 380 times Cos 55 degree = 217.96 m/s V_1y = 380 Sin 55 degree = 311.28 m/s To reach peak height t_1 = V_1y/g = 311.28/9.8 = 31.76 sec Horizontal distance R_1 = V_1n times t_1 = 217.96 times 31.76 = 6923.12M Peak height h_1 = (V_1y)^2/2g = (311.28)^2/2 times 9.8 = 4943.63M Conversion of momentum before and after collision is m_1 V_1x + n_2 V_20 = (m_1 + m_2) times V 14 times 217.96 + m_2 times (-140) = (14 + m_2) times V 3051.44 - 140 m_2 = (14V + m_2) times V horizontal distance R_2 = V times 31.76 R_1 + R_2 = 9.9 km = 9900m So, 6923.12 + 31.76 V = 9900 31.76V = 990 - 6923.12 Therefore V = 93.73m/s \r\n Substitute in the equation 3051.44 + m_2(-140) = (14 + m_2) times 93.73 3051.44 - 140 m_2 = 1312.22 + 93.73 m_2 3051.44 - 1312.22 = 93.73 M_2 + 140 m_2 1739.22 = 233.73 m_2 => m_2 = 1739.22/233.73 => m_2 = 7.44Kg 2 Significant figures = 7.4 kg (Max of 2^nd projectile)