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NISA [10]
3 years ago
8

A bullet of mass 0.010 kg and speed of 200 m/s is brought to rest in a wooden block after penetrating a distance of 0.10 m. The

work done on the bullet by the block is
Physics
1 answer:
irina [24]3 years ago
7 0

Answer:

W = 200 J

Explanation:

Work will be equal to the change in kinetic energy

W = ½mv² - 0

W = ½(0.010)200²

W = 200 J

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Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s
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The initial momentum of the system can be expressed as,

p_i=m_1u_1+m_{2_{}}u_2

The final momentum of the system can be given as,

p_f=m_1v_1+m_{2_{}}v_2

According to conservation of momentum,

p_i=p_f

Plug in the known expressions,

\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}

Thus, the final velocity of second mass is 2.99 m/s.

3 0
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Answer:

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