A solid block of mass m2 = 1.83 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring constant k = 50 N/m whose other end is fixed. Another solid block of mass m1 = 2.03 kg and speed v1 = 1.30 m/s collides with the 1.83 kg block. If the blocks stick together, what is their speed immediately after the collision?
What is the change in kinetic energy in the collision?
What is the maximum compression of the spring?
I was able to do part A and got 0.684 m/s as the speed after the collision but got stuck at B and C, can anyone help?
- NCSLv 77 months agoFavorite Answer
A) just conservation of momentum
2.03kg * 1.30m/s = (2.03 + 1.83)kg * V
V = 0.684 m/s √√√ agreed
B) initial KE = ½ * 2.03kg * (1.30m/s)² = 1.72 J
final KE = ½ * 3.86kg * (0.684m/s)² = 0.902 J
and so the change is -0.813 J
C) KE becomes SPE
0.902 J = ½kx² = ½ * 50N/m * x²
x = 0.19 m
Hope this helps!