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AlekseyPX
3 years ago
10

A professional boxer hits his opponent with a 1035 N horizontal blow that lasts 0.175 s. The opponent's total body mass is 120 k

g and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.(a) The opponent's final velocity after the blow(b) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.
Physics
1 answer:
12345 [234]3 years ago
7 0

Answer:

(a) vf = 1.51 m/s

(b) vf = 36.22 m/s

Explanation:

The rate of change of momentum is equal to the force:

F = \frac{mv_f-mv_i}{t}

Ft = m(v_f-v_i)

where,

F = Force = 1035 N

t = time = 0.175 s

vi = initial speed = 0 m /s

vf = final speed = ?

(a)

m = mass of body = 120 kg

Therefore,

(1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\

<u>vf = 1.51 m/s</u>

<u></u>

(b)

m = mass of head = 5 kg

Therefore,

(1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\

<u>vf = 36.22 m/s</u>

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A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
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Answer:

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Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

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\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

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p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}.

Rewrite p = m\, v to obtain v = (p / m). Since the mass of this cannon is m(\text{cannon}) = 33\; {\rm kg}, the velocity of this cannon would be:

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1 year ago
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Answer:

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