Question 6.6: Two Blocks and a Spring Goal Solve an elastic collision invo...
Two Blocks and a Spring Goal Solve an elastic collision involving spring potential energy.
Problem A block of mass m_{1}=1.60 kg , initially moving to the right with a velocity of +4.00 m/s on a frictionless horizontal track, collides with a massless spring attached to a second block of mass m_{2}=2.10 kg moving to the left with a velocity of -2.50 m/s, as in Figure 6.12a. The spring has a spring constant of 6.00 × 10² N/m. (a) Determine the velocity of block 2 at the instant when block 1 is moving to the right with a velocity of +3.00 m/s, as in Figure 6.12b. (b) Find the compression of the spring.
Strategy We identify the system as the two blocks and the spring. Write down the conservation of momentum equations, and solve for the final velocity of block 2, v_{2 f} . Then use conservation of energy to find the compression of the spring.
(a) Find the velocity v_{2 f} when block 1 has velocity +3.00 m/s.
\begin{aligned}&m_{1} v_{1 i}+m_{2} v_{2 i}=m_{1} v_{1 f}+m_{2} v_{2 f} \\&v_{2 f}=\frac{m_{1} v_{1 i}+m_{2} v_{2 i}-m_{1} v_{1 f}}{m_{2}} \\&=\frac{(1.60 kg )(4.00 m / s )+(2.10 kg )(-2.50 m / s )-(1.60 kg )(3.00 m / s )}{2.10 kg }\end{aligned}
Write the conservation of momentum equation for the system and solve for v_{2 f} :
v_{2 f}=-1.74 m / s
(b) Find the compression of the spring.
Use energy conservation for the system, noticing that potential energy is stored in the spring when it is compressed a distance x:
\begin{aligned}&E_{i}=E_{f} \\&\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}\end{aligned}
Substitute the given values and the result of part (a) into the preceding expression, solving for x.
x=0.173 m
Remarks The initial velocity component of block 2 is -2.50 m/s because the block is moving to the left. The negative value for v_{2 f} means that block 2 is still moving to the left at the instant under consideration.