Question

A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastward at 63.863.8 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.

Answer 1

Answer:  75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7  x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 =  200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

Answer 2

The final speed of the lion–gazelle system immediately after the attack is about 65.9 km/hr

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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 )

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$\large {\boxed {F = \Delta (mv) \div t }$

F = Force ( Newton )

m = Object's Mass ( kg )

v = Velocity of Object ( m/s )

t = Time Taken ( s )

Let us now tackle the problem !

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Given:

mass of lion = m₁ = 169 kg

velocity of lion = v₁ = 77.3 j km/hr

mass of gazelle = m₂ = 31.7 kg

velocity of gazelle = v₂ = 63.8 i km/hr

Asked:

final speed = v = ?

Solution:

We will use Conservation of Momentum to solve the problem as follows:

$p_1 + p_2 = p$

$m_1 v_1 + m_2 v_2 = ( m_1 + m_2 ) \overrightarrow{v}$

$169 ( 77.3 \widehat{j} ) + 31.7 ( 63.8 \widehat{i} ) = ( 169 + 31.7 ) \overrightarrow{v}$

$\overrightarrow{v} \approx 65.1 \widehat{j} + 10.1 \widehat{i}$

$|\overrightarrow{v}| \approx \sqrt{65.1^2 + 10.1^2}$

$|\overrightarrow{v}| \approx 65.9 \texttt{ km/hr}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics