Answered

A 100 g ball moving to the right at 4.0 m/s collides head-on with a 200 g ball that is moving to the left at 3.0 m/s.


If the collision is perfectly elastic, what are the speeds of each ball after the collision?

Verified Answer :

Initial momentum of 100gm ball : (Pi)1 =mv= 100x4= 400 gm m/s = 0.4 kg m/s. Initial momentum of 200gm ball : (Pi)2 = m'v'= 200x(-3) = -600 gm m/s = - 0.6 kg m/s. Therefore total initial momentum is (Pi)1 + (Pi)2 = 0.4 - 0.6 = - 0.2 kg m/s. Final momentum of 100gm ball : (Pf)1 = m(-v1) = - mv1= - 0.1v1 kg m/s. Final momentum of 200gm ball : (Pf)2= mv2= 0.2v2 kg m/s. Therefore total final momentum is (Pf)1 + (Pf)2 = 0.2 v2 - 0.1 v1. Since the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law), linear momentum must be conserved. Therefore, total initial momentum = total final momentum.
0.2 v2 - 0.1 v1 = - 0.2 - - - (i). Now since collision is perfectly elastic(second equation is in pic). Solving i and ii equations simultaneously , we get v1 =0 m/s and v2 = -1 m/s ( minus indicates 200gm ball will continue its motion in left direction with 1m/s). So velocity of 100gm ball is 0 m/s and velocity of 200gm ball is 1m/s in leftward direction.

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johanrusli

The final speed of the 100-g ball is about 5.3 m/s to the left

The final speed of the 200-g ball is about 1.7 m/s to the right

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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 ball 1 = m₁ = 100 g

initial velocity of ball 1 = u₁ = 4.0 m/s

mass of ball 2 = m₂ = 200 g

initial velocity of ball 2 = u₂ = -3.0 m/s

Asked:

final velocity of ball 1 = v₁ = ?

final velocity of ball 2 = v₂ = ?

Solution:

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

\(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\)

\(100(4.0) + 200(-3.0) = 100v_1 + 200v_2\)

\(-200 = 100v_1 + 200v_2\)

\(-2 = v_1 + 2v_2\) → Equation 1

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If the collision is perfectly elastic , then:

\(u_1 - u_2 = v_2 - v_1\)

\(4.0 - (-3.0) = v_2 - v_1\)

\(7.0 = v_2 - v_1\) → Equation 2

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Let's solve the two Equations above:

(Equation 1) + (Equation 2) ↓

\(-2 + 7.0 = (v_1 + 2v_2) + (v_2 - v_1)\)

\(5.0 = 3v_2\)

\(v_2 = 5.0 \div 3\)

\(v_2 = 1\frac{2}{3} \texttt{ m/s}\)

\(v_2 \approx 1.7 \texttt{ m/s}\)

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\(-2 = v_1 + 2v_2\)

\(-2 = v_1 + 2(1\frac{2}{3})\)

\(v_1 = -2 - 3\frac{1}{3}\)

\(v_1 = -5\frac{1}{3} \texttt{ m/s}\)

\(v_1 \approx -5.3 \texttt{ m/s}\)

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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

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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

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