A5.00-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

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

= initial velocity of the ball before collision = 0 m/s

= initial velocity of the stone before collision = 12 m/s

= final velocity of the ball after collision = ?

= final velocity of the stone after collision = - 8.50 m/s

using conservation of momentum

M + m = M + m

(5) (0) + (1.5) (12) = 5 + (1.50) (- 8.50)

= 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

(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

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.

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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:

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

Next , we will use Conservation of Energy as follows:

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

Grade: High School

Subject: Physics

Chapter: Dynamics

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