Question

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The block rests on a frictionless surface. A 600 kg wad of putty is thrown horizontally at the block, hitting it with a speed of 4.4 m/s and sticking. How far does the putty-block system compress the spring?

Answer 1

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    $p_{f}$ = (m1 + m2) $v_{f}$

   p₀ = $p_{f}$

   m1 v₀₁ = (m1 + m2) $v_{f}$

  $v_{f}$ = v₀₁ m1 / (m1 + m2)

   $v_{f}$= 4.4 600 / (600 + 500)

  $v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) $v_{f}$²

After compressing the spring

   $E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = $E_{mf}$

   ½ (m1 + m2) $v_{f}$² = = ½ k x²

   x = $v_{f}$ √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m