Question

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 25.0 N/m. The block rests on a frictionless surface. A 5.70×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.99 m/s and bounced with the same speed of 8.99 m/s in opposite direction. How far does the block compresses the spring?

Answer 1

The total momentum of the block and putty prior to their collision is

(0.454 kg) (0 m/s) + (5.70 × 10⁻² kg) (8.99 m/s) ≈ 0.512 kg•m/s

and the total momentum after the collision is

(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s)

where v is the velocity of the block. Momentum is conserved, so

(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s) = 0.512 kg•m/s

==>   v ≈ 2.26 m/s

The total work done on the block by the spring as it gets compressed by a distance x is equal to the change in the block's kinetic energy:

1/2 (25.0 N/m) x ² = 1/2 (0.454 kg) (2.26 m/s) - 0

==>   x ≈ 0.202 m ≈ 20.2 cm