Question

A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object so that it strikes the stationary smaller object which then slides follows the curved track so that it rises vertically to a maximum height, h_{max}h ​max ​​ . The masses are equipped with elastic bumpers so that the impact between them is an elastic collision. If the larger object has a mass , M = 5.41 ~\text{kg}M=5.41 kg and the smaller object has a mass of m = 1.68~\text{kg}m=1.68 kg, then with what velocity, v_0v ​0 ​​ should the player release the larger object so that the smaller object just reaches the target maximum height of h_{max} = 3.0 mh ​max ​​ =3.0m above the horizontal portion of the track?

Answer 1

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    $Em_{f}$ = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   $Em_{f}$

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      $p_{f}$ = M v1f + m v

   p₀ =   $p_{f}$

   M v₁₀ = M $v_{1f}$+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = $K_{f}$

    ½ M v₁₀² = ½ M $v_{1f}$² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  $v_{1f}$ + m v

    M v1₁₀² = M $v_{1f}$² + m v²

We cleared v1f in the first we replaced in the second

   $v_{1f}$ = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s