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Image for Problem 2 A 1.0 kg cannonball is fired horizontally with a speed of 200 rn/s at two blocks (each with mass 10.

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Answer #1

a) The center mass position is:

x_{cm} = \frac{m_0x_0+m_1x_1+m_2x_2}{m_0 + m_1 + m_2}

x_{cm} = \frac{(1.0)(0)+(10.0)(1)+10(2.0)}{1.0+10.0+10.0}= 1.4286 m

b) We use the conservation of linear mometum:

m_0v_0= (m_0 +m_1)V_1

V_1 = \frac{m_0v_0}{m_0 +m_1} = \frac{(1.0)(200)}{1.0+10.0}= 18.18 m/s

c) we calculate the initial kinetic energy:

K_0 = m_0v_0^2/2 = (1.0)(200)^2/2 = 20000 J

final kinetic energy:

K_f= (m_0 + m_1)V_1^2/2 = (1.0+10.0)(18.18)^2/2 =1817.8 J

d) we have to apply the conservation of linear mometum again:

(m_0 + m_1)V_1 = (m_0 + m_1)v_1 + m_2v_2

using given data we get:

11v_1 + 10v_2 = 200

also we know that this is a elastic collision:

e = -\frac{v_2-v_1}{V_2-V_1} = 1

so:

v_2 - v_1 = 18.18

solving the system we have:

v_1 = 0.867 m/s

v_2 = 19.05 m/s

e) we have:

-M_{cannon}V_{cannon} +m_0v_0 = 0

solving for Vcannon:

V_{cannon} = \frac{m_0v_0}{M_{cannon}} = \frac{1.0(200)}{1000}=0.200 m/s

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