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Novay_Z [31]
3 years ago
14

A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne

arly the same speed with which it hit. The clay sticks to the wall. Which one of these objects experiences the greater momentum change?
Physics
1 answer:
gregori [183]3 years ago
3 0

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

\Delta p = m \Delta v= m (v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

v=-u

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

\Delta p=m(v-u)=m((-u)-u)=-2mu

- For the clay, the final velocity is

v=0

since it sticks to the wall. So, the change in momentum is

\Delta p = m(v-u)=m(0-u)=-mu

So we see that the greater momentum change (in magnitude) is experienced by the ball.

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The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is
Fudgin [204]

Answer:

24cm/s

Explanation:

A=L*w

A'=L'*w'

L=13

w=5

L'=4

w'=6

A=?

A'=?

A=L*w

A=13*5

A=65

A'=L'*w'

A'=4*6

A'=24

*the given lengths are just to throw you off*

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3 years ago
A 5.6 g marble is fired vertically upward using a spring gun. The spring must be compressed 6.4 cm if the marble is to just reac
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Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy

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5 0
2 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
2 years ago
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