Question

Un proyectil de masa m (kg) se dispara con una velocidad v (m/s) contra un bloque de masa 4m inicialmente en reposo. Tras la colisión el sistema proyectil-bloque sale unido como un único cuerpo que se desplaza por una superficie horizontal y rugosa. Sabiendo que el coeficiente de rozamiento de la superficie es μ, determine la distancia máxima que recorre el cuerpo en función de todos los datos del problema.

Answer 1

Answer:

  x = $\frac{v^2}{10 \mu g}$

Explanation:

Let's start the exercise with the definition of a system formed by the projectile and the block, in this case the forces during the collision are internal and the moment is conserved,

initial instant. Just before the crash

         p₀ = m v + 4m 0

final instant. Right after the crash, before the block began to move

        $p_{f}$ = (m + 4m) v_{f}

how the moment is preserved

        p₀ = p_{f}

        m v = 5m v_{f}

       v_{f} = v/5

knowing the speed of the system (projectile + block) we can use the relationship between work and energy

       W = ΔK

starting point. Just when the projectile + block system starts to move

       Em₀ = K = ½ m v_{f}²

final point. When the system is stopped

       Em_{f} = 0

The work of the friction force is

       W = - fr x

the negative sign is because the friction force opposes the motion, let's use Newton's second law to find the friction force

Y Axis  

      N- W = 0

      N = W = mg

The expression for the friction force is

      fr = μ N

     

substituting

     fr = μ mg

      W = - μ mg x

using the energy duty ratio

      - μ mg x = 0 - ½ m $v_{f}^2$

        x =   $\frac{v_{f}^{2} }{2 \mu g}$

we substitute speed

       x = $\frac{v^2}{10 \mu g}$