Question 1: An excellent driver can decelerate at 6.08 m/s2 in an emergency. Calculate the total stopping distance for a car traveling at 11.2 m/s (25 mph)
Student response.
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A. Reaction distance
B. Braking distance
C. Total distance
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The final speed of the 2.3-kg object is about 7.1 m/s

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 object 1 = m₁ = 2.3 kg

initial velocity of object 1 = u₁ = 6.1 m/s

mass of object 2 = m₂ = 3.5 kg

initial velocity of object 2 = u₂ = -4.8 m/s

Asked:

final velocity of object 1 = v₁ = ?

Solution:

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

→ Equation 1

If the collision is perfectly elastic , then:

→ Equation 2

Let's solve the two Equations above:

(Equation 1) - 3.5(Equation 2) ↓

Grade: High School

Subject: Physics

Chapter: Dynamics

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