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Lilit [14]
3 years ago
5

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

ball bounces off with no loss of speed, the clay sticks. What is the change in momentum of the clay and ball, respectively?
O mv; 0
O mv; mv
O 0; mv
O 0, 2mv
O mv; 2 mv
Physics
1 answer:
lana [24]3 years ago
3 0

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

\Delta p=p_2-p_1

The momentum is computed as

p=mv

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

p_1=mv

When it hits the wall, it sticks, thus its final speed is 0 and

p_2=0

The change of momentum is

\Delta p=0-mv=-mv

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

p_2=-mv

The change of momentum is

\Delta p=-mv-mv=-2mv

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

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Sort the sentences based on whether they describe radio waves, visible light waves, or both.
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6 0
3 years ago
Which two changes would decrease the electric force between two charged
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4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p
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Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A ,   n 1  =  4 .2 mol

Number of moles mono atomic gas B ,   n 2  =  3.2mol

Initial energy of gas A ,   K A  =  9500  J

Thermal energy given by gas A to gas B ,   Δ K  =  600 J

Gas constant   R  = 8.314  J / molK

Let  K B  be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0
3 years ago
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