Question 9.7: A Two-Body Collision with a Spring A block of mass m1 = 1.60......

(A) Conceptualize With the help of Figure 9.10a, run an animation of the collision in your mind. Figure 9.10 \mathrm{~b} shows an instant during the collision when the spring is compressed. Eventually, block 1 and the spring will again separate, so the system will look like Figure 9.10a again but with different velocity vectors for the two blocks.

Categorize Because the spring force is conservative, kinetic energy in the system of two blocks and the spring is not transformed to internal energy during the compression of the spring. Ignoring any sound made when the block hits the spring, we can categorize the collision as being elastic and the system as being isolated for both energy and momentum.

Analyze Because momentum of the system is conserved, apply Equation 9.16:

(1) m_{1} v_{1 i}+m_{2} v_{2 i}=m_{1} v_{1 f}+m_{2} v_{2 f}

Because the collision is elastic, apply Equation 9.20:

\text { (2) } v_{1 i}-v_{2 i}=-\left(v_{1 f}-v_{2 f}\right)

Multiply Equation (2) by m_{1} :

(3) m_{1} v_{1 i}-m_{1} v_{2 i}=-m_{1} v_{1 f}+m_{1} v_{2 f}

Add Equations (1) and (3):

2 m_{1} v_{1 i}+\left(m_{2}-m_{1}\right) v_{2 i}=\left(m_{1}+m_{2}\right) v_{2 f}

Solve for v_{2 f} :

v_{2 f}=\frac{2 m_{1} v_{1 i}+\left(m_{2}-m_{1}\right) v_{2 i}}{m_{1}+m_{2}}

Substitute numerical values:

v_{2 f}=\frac{2(1.60 \mathrm{~kg})(4.00 \mathrm{~m} / \mathrm{s})+(2.10 \mathrm{~kg}-1.60 \mathrm{~kg})(-2.50 \mathrm{~m} / \mathrm{s})}{1.60 \mathrm{~kg}+2.10 \mathrm{~kg}}=3.12 \mathrm{~m} / \mathrm{s}

Solve Equation (2) for v_{1 f} and substitute numerical values:

v_{1 f}=v_{2 f}-v_{1 i}+v_{2 i}=3.12 \mathrm{~m} / \mathrm{s}-4.00 \mathrm{~m} / \mathrm{s}+(-2.50 \mathrm{~m} / \mathrm{s})=-3.38 \mathrm{~m} / \mathrm{s}

(B) Conceptualize Focus your attention now on Figure 9.10b, which represents the final configuration of the system for the time interval of interest.

Categorize Because the momentum and mechanical energy of the system of two blocks and the spring are conserved throughout the collision, the collision can be categorized as elastic for any final instant of time. Let us now choose the final instant to be when block 1 is moving with a velocity of +3.00 \mathrm{~m} / \mathrm{s}.

Analyze Apply Equation 9.16:

m_{1} v_{1 i}+m_{2} v_{2 i}=m_{1} v_{1 f}+m_{2} v_{2 f}

Solve for v_{2 f} :

v_{2 f}=\frac{m_{1} v_{1 i}+m_{2} v_{2 i}-m_{1} v_{1 f}}{m_{2}}

Substitute numerical values:

Finalize The negative value for v_{2 f} means that block 2 is still moving to the left at the instant we are considering.

(C) Conceptualize Once again, focus on the configuration of the system shown in Figure 9.10 \mathrm{~b}.

Categorize For the system of the spring and two blocks, no friction or other nonconservative forces act within the system. Therefore, we categorize the system as isolated in terms of energy with no nonconservative forces acting. The system also remains isolated in terms of momentum.

Analyze We choose the initial configuration of the system to be that existing immediately before block 1 strikes the spring and the final configuration to be that when block 1 is moving to the right at 3.00 \mathrm{~m} / \mathrm{s}.

Write a conservation of mechanical energy equation for the system:

K_{i}+U_{i}=K_{f}+U_{f}

Evaluate the energies, recognizing that two objects in the system have kinetic energy and that the potential energy is elastic:

\frac{1}{2} m_{1} v_{1 i}^{2}+\frac{1}{2} m_{2} v_{2 i}^{2}+0=\frac{1}{2} m_{1} v_{1 f}^{2}+\frac{1}{2} m_{2} v_{2 f}^{2}+\frac{1}{2} k x^{2}

Substitute the known values and the result of part (B) :

Finalize This answer is not the maximum compression of the spring because the two blocks are still moving toward each other at the instant shown in Figure 9.10 b. Can you determine the maximum compression of the spring?