In the collision pictured in Figure 8.10, two carts collide inelastically. Cart 1 (denoted m1 carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and

Question

In the collision pictured in Figure 8.10, two carts collide inelastically. Cart 1 (denoted $m_1$ carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of 2.00 m/s . Cart 2 (denoted $m_2$ in Figure 8.10) has a mass of 0.500 kg and an initial velocity of −0.500 m/s . After the collision, cart 1 is observed to recoil with a velocity of −4.00 m/s . (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)?

Strategy
We can use conservation of momentum to find the final velocity of cart 2, because $F_{net} = 0$ (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring.

Strategy
We can use conservation of momentum to find the final velocity of cart 2, because [latex] F_{net} = 0 [/latex] (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring.

Accepted Answer

Solution for (a)
As before, the equation for conservation of momentum in a two-object system is

[latex] m_1 v_1 + m_2v_2 = m_1v^′_1 + m_2v^′_2 . [/latex] (8.53)

The only unknown in this equation is [latex] v^′_2 [/latex] . Solving for [latex] v^′_2 [/latex] and substituting known values into the previous equation yields

[latex] v^′_2 = \frac{m_1 v_1 + m_2v_2 − m_1 v^′_1}{m_2} [/latex] (8.54)

[latex] = \frac{(0.350 kg)(2.00 m/s) + (0.500 kg)(−0.500 m/s)}{0.500 kg} - \frac{(0.350 kg)(−4.00 m/s)}{0.500 kg} [/latex]

= 3.70 m/s.

Solution for (b)
The internal kinetic energy before the collision is

[latex] KE_{int} =\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 [/latex] (8.55)

[latex] = \frac{1}{2} (0.350 kg)(2.00 m/s)^2 +\frac{1}{2} (0.500 kg)( – 0.500 m/s)^2 [/latex]

= 0.763 J.

After the collision, the internal kinetic energy is

[latex] KE^′_{int} =\frac{1}{2} m_1 v_1^{′2} + \frac{1}{2} m_2 v_2^{′2} [/latex] (8.57)

[latex] = \frac{1}{2} (0.350 kg)(-4.00 m/s)^2 +\frac{1}{2} (0.500 kg)(3.70 m/s)^2 [/latex]

= 6.22 J.

The change in internal kinetic energy is thus

[latex] KE′_{int} − KE_{int} = 6.22 J − 0.763 J [/latex]

= 5.46 J.

Discussion

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.

Question 8.6: In the collision pictured in Figure 8.10, two carts collide ...

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