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azamat
3 years ago
6

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi

nt with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.
A. Determine the pre-collision momentum of the ball.
B. Determine the post-collision momentum of the ball.
C. Determine the momentum change of the ball.
D. Determine the velocity change of the racket.
Physics
1 answer:
Kipish [7]3 years ago
8 0

Answer:

A) pbin = 1.535 Kgm/s (+)

B) pbf = 1.696 Kgm/s (-)

C) Δp = 3.3925 Kgm/s

D) Δvr = 10.249 m/s

Explanation:

Given

Mass of the ball: m = 57.5 g = 0.0575 Kg

Initial speed of the ball: vbi = 26.7 m/s

Mass of the racket: M = 331 g = 0.331 Kg

Final speed of the ball: vbf = 29.5 m/s

A) We use the formula

pbin = m*vbi = 0.0575 Kg*26.7 m/s = 1.535 Kgm/s (+)

B) pbf = m*vbf = 0.0575 Kg*29.5 m/s = 1.696 Kgm/s (-)

C) We use the equation

Δp = pbf - pbin = 1.696 Kgm/s - (-1.535 Kgm/s) = 3.3925 Kgm/s

D) Knowing that

Δp = 3.3925 Kgm/s

we can say that

Δp = M*Δvr

⇒  Δvr = Δp / M

⇒  Δvr = 3.3925 Kgm/s / 0.331 Kg

⇒  Δvr = 10.249 m/s

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Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

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a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

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v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

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Answer:

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