Question

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. what is the magnitude of the change in momentum of the stone?

Answer 1

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

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Further explanation

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

where:

I = impulse on the object ( kg m/s )

∑F = net force acting on object ( kg m /s² = Newton )

t = elapsed time ( s )

Let us now tackle the problem!

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Given:

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

Asked:

magnitude of the change of momentum of the stone = Δp = ?

Solution:

Firstly, we will calculate the final speed of the ball as follows:

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

$v = - v_o \sqrt{70 \%}$ → negative sign due to ball rebounds

$v = - v_o \sqrt{0.7} \texttt{ m/s}$

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Next, we could find the magnitude of the change of momentum of the stone as follows:

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

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Learn more

• Velocity of Runner : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302
• Average Speed of Plane : brainly.com/question/12826372
• Impulse : brainly.com/question/12855855
• Gravity : brainly.com/question/1724648

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics