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Marina86 [1]
2 years ago
6

Two particles collide and stick together. If no external forces act on the two particles, which of the following is correct for

the change in total momentum Δp and the change in total kinetic energy ΔK of the two particles?
a. p<0, K<0.
b. p<0, K=0.
c. p=0, K<0.
d. p=0, K=0.
e. p=0, K>0.
Physics
1 answer:
Semmy [17]2 years ago
4 0

Answer:

c.\Delta p=0 and \Delta k

Explanation:

We are given that two particles collide  and stick together.

If there is no external force act on the two particles then ,it is inelastic collision.

Inelastic collision: There is some loss of kinetic energy but the momentum is conserved.

According to law of conservation of momentum

Initial momentum=Final momentum

Change in momentum=Final momentum-Initial momentum=0

Change in momentum=\Delta p=0

Initial kinetic energy is greater than final kinetic energy.

Change in kinetic energy=Final kinetic energy-kinetic energy=- negative

\Delta k

Hence, option c is true.

c.\Delta p=0 and \Delta k

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They hit at the same time

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2 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
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A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".
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Answer:

Part a)

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Part b)

v = 2.28 m/s

Part c)

v = 177.66 m/s

Part d)

W = 1012.7 J

Part e)

v = 2.1 m/s

Part f)

E = 1037.2 J

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

\theta = 38^o

so the maximum height reached by the pendulum is given as

h = L(1 - cos\theta)

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U = mgh

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U = 13 J

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

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Part c)

now by momentum conservation we can say

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Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

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Part f)

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