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Two carts mounted on an air track are moving toward one another. Cart 1 has a speed of 0.8 m/s and a mass of 0.43 kg. Cart 2 has a mass of 0.61 kg.(a) If the total momentum of the system is to be zero, what is the initial speed of Cart 2 in meters per second?m/sASK YOUR TEACHER(b) Does it follow that the kinetic energy of the system is also zero since the momentum of the system is zero?YesO No(c) Determine the system's kinetic energy in joules in order to substantiate your answer to part (b).

Two carts mounted on an air track are moving toward one another. Cart 1 has a speed-example-1
User Faran Khan
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Given:

mass of cart 1, m1 = 0.43 kg

Speed of cart 1, v1 = 0.8 m/s

Mass of cart 2, m2 = 0.61 kg

Let's solve for the following:

• (a). If the total momentum is zero. Let's find the initial speed of cart 2.

Since the total momentum is zero, apply the law of conservation of momentum.


m_1v_1-m_2v_2=0

Where v2 is the initial speed of cart 2.

Thus, we have:


\begin{gathered} m_2v_2=m_1v_1 \\ \\ v_2=(m_1v_1)/(m_2) \end{gathered}

Plug in the values in the equation and solve for v2:


\begin{gathered} v_2=(0.43*0.8)/(0.61) \\ \\ v_2=(0.344)/(0.61) \\ \\ v_2=0.56\text{ m/s} \end{gathered}

Therefore, the speed of cart 2 is 0.56 m/s.

• (b), Does it follow that the kinetic energy of the system is also zero since the momentum of the system is zero?

The kinetic energy if the system is given by:


KE=(1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2

The kinetic energy of the system must not be zero because the momentum of the system is zero.

No, it does NOT follow that the kinetic energy of the system is also zero since the momentum of the system is zero.

• (c). Determine the system's kinetic energy in joules in order to substantiate your answer to part (b).

Apply the formula for KE used in part B:


\begin{gathered} KE=(1)/(2)m_1v_1^2+(1)/(2)m2v2^2 \\ \\ KE=((1)/(2)*0.43*0.8)+((1)/(2)*0.61*0.56) \\ \\ KE=0.172+0.1708 \\ \\ KE=0.3428\text{ J} \end{gathered}

Therefore, the kinetic energy is 0.34 Joules.

ANSWER:

(a). 0.56 m/s

(b). No

(c). 0.34 J

User James Heazlewood
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