Question

use the worked example above to help you solve this problem. a block of mass ( m_1 = 1.75 ) kg, initially moving to the right with a velocity of ( +3.77 ) m/s on a frictionless horizontal track, collides with a massless spring attached to a second block of mass ( m_2 = 2.44 ) kg moving to the left with a velocity of ( -2.82 ) m/s, as shown in figure a. the spring has a spring constant of ( 5.88\times10^{2} ) 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 shown in figure b. (indicate the direction with the sign of your answer.)
\text{m/s}
(b) find the compression of the spring.
\text{m}
exercise
hints: getting started | im stuck!
use the values from practice it to help you work this exercise. consider the instant that block 2 is at rest.
(a) find the velocity of block 1. (indicate the direction with the sign of your answer.)
v_{1f}=\text{m/s}
(b) find the compression of the spring.
x=\text{m}

Explanation:

Part (a)

Step1: Apply conservation of momentum

The law of conservation of momentum states that \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\).
We are given \(m_1 = 1.75\space kg\), \(v_{1i}=+ 3.77\space m/s\), \(m_2 = 2.44\space kg\), \(v_{2i}=-2.82\space m/s\) and \(v_{1f}=+3.00\space m/s\).
Substitute the values into the momentum - conservation equation:
\(1.75\times3.77+2.44\times(- 2.82)=1.75\times3.00 + 2.44\times v_{2f}\)

Step2: Solve for \(v_{2f}\)

First, calculate the left - hand side:
\(1.75\times3.77+2.44\times(-2.82)=6.5975-6.8808=-0.2833\)
The right - hand side is \(1.75\times3.00 + 2.44\times v_{2f}=5.25+2.44v_{2f}\)
Set them equal: \(-0.2833 = 5.25+2.44v_{2f}\)
\(2.44v_{2f}=-0.2833 - 5.25=-5.5333\)
\(v_{2f}=\frac{-5.5333}{2.44}\approx - 2.27\space m/s\)

Part (b)

Step1: Use conservation of mechanical energy

The initial kinetic energy \(K_i=\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2\)
\(K_i=\frac{1}{2}\times1.75\times(3.77)^2+\frac{1}{2}\times2.44\times(-2.82)^2\)
\(K_i=\frac{1}{2}\times1.75\times14.1129+\frac{1}{2}\times2.44\times7.9524\)
\(K_i = 12.3005+9.7029=22.0034\space J\)
The final kinetic energy \(K_f=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2\)
\(K_f=\frac{1}{2}\times1.75\times(3.00)^2+\frac{1}{2}\times2.44\times(-2.27)^2\)
\(K_f=\frac{1}{2}\times1.75\times9+\frac{1}{2}\times2.44\times5.1529\)
\(K_f = 7.875+6.2937=14.1687\space J\)
The potential energy of the spring \(U=\frac{1}{2}kx^2\), where \(k = 5.88\times10^{2}\space N/m\)
By conservation of energy \(K_i=K_f+\frac{1}{2}kx^2\)
\(\frac{1}{2}kx^2=K_i - K_f\)
\(x^2=\frac{2(K_i - K_f)}{k}\)
Substitute \(K_i - K_f=22.0034 - 14.1687 = 7.8347\space J\) and \(k = 588\space N/m\)
\(x^2=\frac{2\times7.8347}{588}\)
\(x^2=\frac{15.6694}{588}\approx0.02665\)
\(x=\sqrt{0.02665}\approx0.163\space m\)

Exercise Part (a)

Step1: Apply conservation of momentum

Using \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\), with \(v_{2f} = 0\)
\(1.75\times3.77+2.44\times(-2.82)=1.75\times v_{1f}+2.44\times0\)
\(6.5975-6.8808 = 1.75v_{1f}\)
\(v_{1f}=\frac{6.5975 - 6.8808}{1.75}=\frac{-0.2833}{1.75}\approx - 0.162\space m/s\)

Exercise Part (b)

Answer:

(a) \(-2.27\space m/s\)
(b) \(0.163\space m\)
Exercise (a) \(-0.162\space m/s\)
Exercise (b) \(0.273\space m\)