COE: T_1 + V_1 - T_2 + V_2 T_1 = 1/2 m_AV^2_A_1 + 1/2 m_BV^2_B_1 + 1/2 m_CV^2_C_

Question

COE: T_1 + V_1 - T_2 + V_2 T_1 = 1/2 m_AV^2_A_1 + 1/2 m_BV^2_B_1 + 1/2 m_CV^2_C_1 V_1 = m_Ag delta y_A + m_Bg delta y_B + m_Cg delta y_C COM: m_Av_A_1 + m_Bv_b_1 + m_Cv_C_1 = m_Av_A_2 + m_Bv_B_2 + m_Cv_C_2 A 2 kg cart (B) with a 1 kg mass (C) that is suspended from it by a 3 m long wire, is initially at rest when it is struck by a 0.03 kg bullet (A) that is travelling at 450 m/s. Assuming the bullet lodges into the cart at impact, determine: A) The initial velocity of the cart after impact B) The velocity of the mass at its maximum height C) The angle, theta, of the wire with the vertical at the mass's maximum height

Explanation / Answer

(A) Applying momentum conservation,

0.03 x 450 = (2 + 0.03) v

v = 6.65 m/s

(B) at maximum height,

cart and mass will travel with same velocity.

(2 + 0.03) (6.65) = (1 + 2 + 0.03) v

v = 4.45 m/s

(C) Applying energy conservation,

(1 x 9.8 x h) + (1 + 2+ 0.03) (4.45^2) /2 = (2 + 0.03) (6.65^2) /2

9.8h + 20.10 = 44.86

h = 2.53 m

h = L (1 - cos(theta))

2.53 / 3 = 1 - cos(theta)

cos(theta) = 0.157

theta = 81 deg

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