# Physics question springs?

A 0.500 kg block is sitting on a horizontal, frictionless surface. The block is connected to a horizontal spring with a force constant of 124 N/m. The other end of the horizontal spring rests against a wall. When a 100.0 g arrow is fired into the wooden block, the spring compresses by 25 cm.

(a) What is the maximum potential energy of the spring?

(b) What was the speed of the arrow and block just after collision?

(c) What was the initial kinetic energy of the arrow?

(d) Explain any difference between (a) and (c).

### 1 Answer

- AshLv 72 months agoFavorite Answer
(a) Spring PE, U = ½kx² = ½(124)(0.25)² = 3.875 J

(b) KE of block+bullet = Spring PE

½(m₁+m₂)v² = 3.875

v² = 2(3.875)/(m₁+m₂)

v² = 7.75/(0.1000+0.500)

v² = 7.75/0.600

v = 3.59 m/s ← speed of the arrow and block just after collision

(c) For non-elastic collision

m₁u₁ + m₂u₂ = (m₁+m₂)v

subscript 1 for bullet and subscript 2 for block

since u₂ = 0

m₁u₁ = (m₁+m₂)v

u₁ = (m₁+ m₂)v/m₁

u₁ = (0.1000 + 0.500)3.59/(0.1000)

u₁ = 21.5 m/s

KE of arrow = ½m₁u₁² = ½(0.1000)(21.5)² = 23.1 J

d) Difference in energy = 23.1 - 3.875 = 19.2 J ← energy lost in collision of arrow qith the block