Answer:
The linebacker was moving at [tex]4.375 \,\,\frac{m}{s}[/tex]
Explanation:
We use conservation of momentum to solve this problem.
The initial momentum of the system of two players is the addition of the two linear momentum of each individual player. Recalling that the liner momentum P is defined as the product of the mass of the object (in our case the mass of each player) times its velocity, and noticing that the quarterback (of mass 84 kg) was at rest before the collision, we can write the initial momentum [tex]P_i[/tex] of the system as:
[tex]P_I=(112\,kg)\,v_i + (84\,kg)\,*0\\P_i=(112\,kg)\,v_i[/tex]
where the quantity [tex]v_i[/tex] represents the unknown initial velocity of the linebacker of mass 112 kg.
We can also write the expression for the final linear momentum of the system of two players moving together a a speed of 2.5 m/s:
[tex]P_f=(112+84)\,*(2.5\,m/s)\\P_f=(196\, kg)\,kg\,*(2.5\,m/s)\\P_f=490\,kg\,\,m/s[/tex]
Since the total momentum is conserved in collisions, we set an equation making the initial momentum equal the final momentum of the system of two players. We then solve for the unknown initial velicity:
[tex]P_i=P_f\\(112\,kg)\,v_i=490\,kg\,\,\frac{m}{s} \\v_i=\frac{490}{112} \,\,\frac{m}{s}\\v_i=4.375 \,\,\frac{m}{s}[/tex]
The linebacker moving just before the collision at the speed of:
"4.375 m/s".
According to the question,
Mass, m₁ = 112.0 kg
m₂ = 84.0 kg
Speed = 2.50 m
By using Conservation of momentum,
→ [tex]P_I[/tex] = m₁ v₁ + m₂ v₂
By substituting the values,
= 112 v₁ + 84 × 0
= 112 v₁
Now, the final linear momentum,
→ [tex]P_f[/tex] = (112 + 84) × 2.5
= 196 × 2.5
= 490 kg.m/s
hence,
The speed will be:
→ [tex]P_I[/tex] = [tex]P_f[/tex]
112 v₁ = 490 kg.m/s
v₁ = [tex]\frac{490}{112}[/tex]
= 4.375 m/s
Thus the above answer is correct.
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