If m_1 = 2.0 kg, m_2 = 3.0 kg, and m_3 = 4.0 kg, find the x and y coordinates of

Question

If m_1 = 2.0 kg, m_2 = 3.0 kg, and m_3 = 4.0 kg, find the x and y coordinates of the center of mass of this system. Ballistic Pendulum: A bullet with mass fired into a block of wood with mass M suspended like a pendulum and makes a completely inelastic collision with it. After the impact of the bullet, the block swings up to maximum height h. Given the values of h = 5.00 cm m, m = 6.75 g = 0.00675 kg, and M = 2.50 kg, (a) What is the initial velocity v_x of the bullet in m/s? (b) What is the initial velocity V_x of the bullet/block system right after impact? (c) What is the energy of the bullet right before kinetic right after impact? (d) Where is the kinetic energy of the bullet/block system right after impact? Where did the initial energy of the bullet go? (Remember Use Conservation of Linear Momentum to analyze the impact. Then, use conservation of Total Mechanical Energy to see how high the bullet/block system swings.) A stone of mass m_A = 18 kg initially slides on a frictionless surface at 2.0 m/s in the positive direction. It collides with another stone of mass m_B = 14 kg which is initially at rest. After the stone moving 1.2 m/s in a direction that makes an angle alpha = 32 degree with respect to the positive x-direction. What is the makes an angle alpha final velocity of stone B?

Explanation / Answer

The X and Y coordinates of center of mass of the given system can be calculated as follows:

Xcom = m1x1 + m2x2 + m3x3/(m1 + m2 + m3)

Xcom = 2 x 0 + 3 x 2 + 4 x 1 / (2 + 3 + 4) = 1.11

Ycom = m1y1 + m2y2 + m3y3/(m1 + m2 + m3)

Ycom = 2 x 0 + 3 x 1 + 4 x 2 / (2 + 3 + 4) = 1.22

Hence, (Xcom,Ycom) = (1.11,1.22)m

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