Respuesta :
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
[tex] V_{bi} [/tex] = initial velocity of the ball before collision = 0 m/s
[tex] V_{si} [/tex] = initial velocity of the stone before collision = 12 m/s
[tex] V_{bf} [/tex] = final velocity of the ball after collision = ?
[tex] V_{sf} [/tex] = final velocity of the stone after collision = - 8.50 m/s
using conservation of momentum
M[tex] V_{bi} [/tex] + m[tex] V_{si} [/tex] = M[tex] V_{bf} [/tex] + m[tex] V_{sf} [/tex]
(5) (0) + (1.5) (12) = 5 [tex] V_{bf} [/tex] + (1.50) (- 8.50)
[tex] V_{bf} [/tex] = 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[tex] V_{bf}^{2} [/tex]
(9.8) h = (0.5) (6.15)²
h = 1.93 m
The maximum height that is reached by the ball is about 1.93 m
[tex]\texttt{ }[/tex]
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.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
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:
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
[tex]5.00(0) + 1.50(12.0) = 5.00v_1 + 1.50(-8.50)[/tex]
[tex]0 + 18 = 5v_1 - 12.75[/tex]
[tex]5v_1 = 18 + 12.75[/tex]
[tex]5v_1 = 30.75[/tex]
[tex]v_1 = 30.75 \div 5[/tex]
[tex]v_1 = 6.15 \texttt{ m/s}[/tex]
The speed of the ball is 6.15 m/s just after it hit by the stone.
[tex]\texttt{ }[/tex]
Next , we will use Conservation of Energy as follows:
[tex]Ek = Ep[/tex]
[tex]\frac{1}{2}m_1(v_1)^2 = m_1 g h[/tex]
[tex]\frac{1}{2}(v_1)^2 = g h[/tex]
[tex]h = \frac{1}{2}(v_1)^2 \div g[/tex]
[tex]h = \frac{1}{2}(6.15)^2 \div 9.8[/tex]
[tex]h \approx 1.93 \texttt{ m}[/tex]
The maximum height that is reached by the ball is about 1.93 m
[tex]\texttt{ }[/tex]
Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
- Newton's Law of Motion: https://brainly.com/question/10431582
- Example of Newton's Law: https://brainly.com/question/498822
[tex]\texttt{ }[/tex]
Answer details
Grade: High School
Subject: Physics
Chapter: Dynamics
[tex]\texttt{ }[/tex]
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
