Question

3. High speed strobe light photographs show the head of a 215 g golf club traveling at 55 m/s just before it strikes a 46
golf ball at rest on a tee. After the collision, the club travels in the same direction at 42 m/s. Use the law of conservati
of momentum to find the speed of the golf ball just after impact.

Answer 1

Answer:

61 m/s

Explanation:

Momentum before collision = momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(215 g) (55 m/s) + (46 g) (0 m/s) = (215 g) (42 m/s) + (46 g) v

v ≈ 61 m/s

Answer 2

the conservation of the momentum allows to find the the result for the speed of the golf ball after the stroke is:

        v = 60.8 m / s

Given parameters

• The mass of the golf club M = 215 g = 0.215 kg

• The initial velocity v₁ = 55 m / s

• The final velocity v₂ = 42 m / s

• Golf ball mass m = 46 g = 0.046 kg

To find

• The speed of the golf ball after hit.

Momentum is defined by the product of mass and velocity.

        p = m v

Where me is the mass and v the velocity of the body.

In an isolated system the forces are internal, therefore the momentum is conserved.

Let's write the momentum in two moments:

Initial instant. Before hit.

        p₀ = M v₁ + 0

Final instatne. After the coup.

        p_f = M v₂ + m v

If we define the system as formed by the golf club and the golf ball, this system is isolated and the momentum is conserved.

         p₀ = p_f

         M v₁ = M v₂ + m v

         v = $\frac{M}{m} ( v_1 - v_2)$  

Let's calculate

         v = $\frac{0.215}{0.046} \ (55-42)$  

         v = 60.75 m / s

In conclusion, using the conservation of momentum, we can find tfor the speed of the golf ball after the stroke is:

        v = 60.8 m / s

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