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hichkok12 [17]
3 years ago
11

Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i

nitially traveling away from Aat 2.0m/s. After the collision the center of mass of the two cart system has a speed of:____________.A. zeroB. 0.33m/sC. 2.3m/sD. 2.5m/sE. 5.0m/s

Physics
1 answer:
inysia [295]3 years ago
3 0

Answer:

\large \boxed{\text{C. 2.3 m/s}}

Explanation:

Data:

m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\

Calculation:

This is a perfectly inelastic collision.  The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}}  + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}

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A bullet of mass 0.016 kg traveling horizontally at a speed of 280 m/s embeds itself in a block of mass 3 kg that is sitting at
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The conservation of momentum states that the total initial momentum is equal to the total final momentum:

p_i = p_f\\m u_b + M u_B = (m+M)v

where

m = 0.016 kg is the mass of the bullet

u_b = 280 m/s is the initial velocity of the bullet

M = 3 kg is the mass of the block

u_B = 0 is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s

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The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J

(c) The Energy Principle isn't valid for an inelastic collision.

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