Question

A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 200 g glider is pushed toward it from the far end of the track at a speed of 170 cm/s . It collides with and sticks to the 750 g glider.What are the amplitude and period of the subsequent oscillations?

Answer 1

Answer:

the amplitude of  the subsequent oscillations is 0.11  m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider,  $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of  the subsequent oscillations is A = 0.11  m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and  $v_{f}$ is the final speed.

to find $v_{f}$, we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

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

$v_{1}$ = 0, thus

(0.75+0.2)$v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11  m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π$\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π$\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s