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# Physics Question?

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?

Part B

What is the change in kinetic energy in the collision?

Part C

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?

### 1 Answer

- 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!