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Troyanec [42]
3 years ago
8

High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46

.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.
Physics
1 answer:
MissTica3 years ago
6 0

Answer:

The speed of the golf ball just after the impact is 73.04 m/s.

Explanation:

Given that,

The mass of golf club, m₁ = 183 g = 0.183 kg

The mass of golf ball, m₂ = 46.6 g = 0.0466 kg

The initial speed of golf club, u₁ = 58.6 m/s

The initial speed of a golf ball, u₂ = 0

The final speeds of club, v₁ = 40 m/s

We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s

So, the speed of the golf ball just after the impact is 73.04 m/s.

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Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

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The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

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x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

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Answer:

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