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Solnce55 [7]
2 years ago
14

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45

.9 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 30.7 m/s. Find the speed of the golf ball immediately after impact. Answer in units of m/s.
Physics
1 answer:
Helga [31]2 years ago
8 0

Answer:

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

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A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 2
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Answer:

0.08m/s

Explanation:

Given data

M1= 69kg

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M2= 0.22kg

v2= 25.2m/s

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Total mass of skater + snowball = 69+ 0.22 = 69.22kg

Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s

After snowball is thrown:

Let's call the velocity of the skater V.

Total momentum = momentum of skater + momentum of snowball

=69.22V + (5.544)

= 69.22V + 5.544

So:

180.7  = 69.22V+5.544

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175.156= 69.22V

V= 175.156/69.22

V = 2.53m/s

The total momentum after catching the snowball is mV or:

(69.0 + 0.22) x V

So:

5.544= 69.22V

V= 5.544/69.22

V=0.08m/s

The velocity of the ice skater after throwing the snowball is 0.08m/s

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