A mass m1 = 7.65 kg, is at rest on a frictionless horizontal surface and connected to a wall by a spring with k = 67.9 N/m, as shown in the figure. A second mass, m2 = 5.29 kg, is moving to the right at v0 = 15.3 m/s. The two masses collide and stick together.

Question

the figure is
htt p://s2.dosya.tc/server22/DIDqtx/P043figure.png. html

a) What is the maximum compression of the spring?
(in m) 2.731 m

b) How long will it take after the collision to reach this maximum compression?
(in s) 0.6857 s

i tried everything ,just found out the second one but still have a problem about the first one.

Answer 1

First find the speed after collision from conservation of momentum
initial momentum:
m v = 5.29 * 15.3 = 80.9 which will be momentum after crash

after crash
m = 5.29+7.65 = 12.94 kg
so for v after crash:
12.94 v = 80.9
v = 6.25 m/s after crash
v at end is zero and energy after crash is conserved so
(1/2)(12.94)(6.25)^2 = (1/2) (67.9) x^2
x^2 = 7.45
and
x = 2.73 meters I agree

time = 1/4 of period T
T = 2 pi sqrt(m/k)
= 2 pi sqrt (12.94/67.9)
= 2.74 seconds
1/4 T = .6857 seconds sure enough

Answer 2

thank you!