Question

A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally to the right at 12.0 m/s. the stone rebounds to the left with a speed of 8.50 m/s, and the ball swings to a maximum height h above its original level. the value of h is closest to

Answer 1

consider the right direction as positive and left direction as negative.

M = mass of the ball = 5 kg

m = mass of stone = 1.50 kg

$V_{bi}$ = initial velocity of the ball before collision = 0 m/s

$V_{si}$ = initial velocity of the stone before collision = 12 m/s

$V_{bf}$ = final velocity of the ball after collision = ?

$V_{sf}$ = final velocity of the stone after collision = - 8.50 m/s

using conservation of momentum

M$V_{bi}$ + m$V_{si}$ = M$V_{bf}$ + m$V_{sf}$

(5) (0) + (1.5) (12) = 5 $V_{bf}$ + (1.50) (- 8.50)

$V_{bf}$ = 6.15 m/s

h = height gained by the ball

using conservation of energy

Potential energy gained by ball at Top = kinetic energy at the bottom

Mgh = (0.5) M$V_{bf}^{2}$

(9.8) h = (0.5) (6.15)²

h = 1.93 m

Answer 2

The maximum height that is reached by the ball is about 1.93 m

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of the ball = m₁ = 5.00 kg

mass of the stone = m₂ = 1.50 kg

initial velocity of the stone = u = 12.0 m/s

final velocity of the stone = v = -8.50 m/s

Asked:

maximum height of the ball = h = ?

Solution:

Firstly, we will use Conservation of Momentum Law as follows:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

$5.00(0) + 1.50(12.0) = 5.00v_1 + 1.50(-8.50)$

$0 + 18 = 5v_1 - 12.75$

$5v_1 = 18 + 12.75$

$5v_1 = 30.75$

$v_1 = 30.75 \div 5$

$v_1 = 6.15 \texttt{ m/s}$

The speed of the ball is 6.15 m/s just after it hit by the stone.

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Next , we will use Conservation of Energy as follows:

$Ek = Ep$

$\frac{1}{2}m_1(v_1)^2 = m_1 g h$

$\frac{1}{2}(v_1)^2 = g h$

$h = \frac{1}{2}(v_1)^2 \div g$

$h = \frac{1}{2}(6.15)^2 \div 9.8$

$h \approx 1.93 \texttt{ m}$

The maximum height that is reached by the ball is about 1.93 m

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant