Physics: During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 3.8 m/s, grabs and holds onto him so that they move off together with a velocity of 2.3 m/s. If the mass of the tackler is 130 kg, determine the mass of the receiver in kilograms. Assume momentum is conserved.

Given,The initial position of the boat, d_i=(5i-2j+4k) mThe initial velocity of the boat, u=(4i+j-3k) m/sThe time period, t=8 sThe final velocity of the boat, v=(12i-7j+13k) m/sTo find:a) The position of the object after 8 sb) The acceleration of the object.Explanation:a)From the equation of motion, the total distance traveled by the object is given by,On substituting the known values,Thus the final position of the boat is,b)The acceleration of an object is given by,On substituting the known values,Final...

During a football game, a receiver has just caught

Ответ:

salam4704

22.01.2022

Physics

The mass of the receiver is 85 kg.

Explanation:

Given:

Mass of the tackler (M) = 130 kg

Mass of the receiver = 'm' (Assume)

Initial velocity of the receiver (u) = 0 m/s

Initial velocity of the tackler (U) = 3.8 m/s

Final combined velocity (v) = 2.3 m/s

So, as per question, the momentum is conserved. Therefore, the total initial momentum is equal to the total final momentum.

Total initial momentum is given as:

$P_i=mu+MU\\\\P_i=0+130\times 3.8\\\\P_i=494\ kg\cdot m/s$

Total final momentum is given as:

$P_f=(m+M)v\\\\P_f=(m+130)2.3$

Now, as per momentum conservation:

$P_f=P_i\\\\(m+130)2.3=494\\\\m+130=\frac{494}{2.3}\\\\m=214.78-130\\\\m=84.78\ kg\approx85\ kg$

Therefore, the mass of the receiver is 85 kg.

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Ответ:

moneykingmarco079

12.11.2022

Physics

Given,

The initial position of the boat, d_i=(5i-2j+4k) m

The initial velocity of the boat, u=(4i+j-3k) m/s

The time period, t=8 s

The final velocity of the boat, v=(12i-7j+13k) m/s

To find:

a) The position of the object after 8 s

b) The acceleration of the object.

Explanation:

From the equation of motion, the total distance traveled by the object is given by,

$d=\frac{1}{2}(v+u)t$

On substituting the known values,

$\begin{gathered} d=\frac{1}{2}[(12\hat{i}-7\hat{j}+13\hat{k)}+(4\hat{i}+\hat{j}-3\hat{k})]8 \\ =4\times(16\hat{i}-6\hat{j}+10\hat{k}) \\ =(64\hat{i}-24\hat{j}+40\hat{k})\text{ m} \end{gathered}$

Thus the final position of the boat is,

$\begin{gathered} d_f=d_i+d \\ =\left(5\hat{i}-2\hat{j}+4\hat{k}\right)+(64\hat{i}-24\hat{j}+40\hat{k}) \\ =(69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{gathered}$

The acceleration of an object is given by,

$a=\frac{v-u}{t}$

On substituting the known values,

$\begin{gathered} a=\frac{(12\hat{i}-7\hat{j}+13\hat{k)}-(4\hat{i}+\hat{j}-3\hat{k})}{8} \\ =\frac{8\hat{i}-8\hat{j}+16\hat{k}}{8} \\ =(\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{gathered}$

Final

a) Thus the position of the boat after 8 s is

$\begin{equation*} (69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{equation*}$

The acceleration of the boat is

$\begin{equation*} (\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{equation*}$

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