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In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. you throw the ball at a speed of 3 m/s. when the ball impacts the target, it sticks to it and they drift off together. 1) how much energy is in the translational energy of the block+ball system after the collision?

Respuesta :

skyluke89
It is not specified but I assume the target is initially at rest.

We can use the conservation of momentum to calculate the final velocity of the ball+block system. The initial momentum is the one of the ball, since the target is at rest:
[tex]p_i = m_b v_b[/tex]
where [tex]m_b = 0.1 kg[/tex] is the mass of the ball and [tex]v_b = 3 m/s[/tex] is the initial velocity of the ball.

The final momentum is instead:
[tex]p_f = (m_b + m_B)v_f[/tex]
where [tex]m_B=1 kg[/tex] is the mass of the block. For the conservation of momentum,
[tex]p_i=p_f[/tex]
so we can find the final velocity of the ball+block system:
[tex]m_B v_b = (m_b+m_B)v_f[/tex]
[tex]v_f = \frac{m_b v_b}{m_b+m_B} = \frac{(0.1 kg)(3 m/s)}{0.1 kg+1 kg} =0.27 m/s[/tex]

And the translational energy of the block+ball system after the collision is equal to its kinetic energy:
[tex]K= \frac{1}{2}(m_b+m_B)v_f^2= \frac{1}{2}(0.1 kg+1 kg)(0.27 m/s)^2=0.04 J [/tex]

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