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kodGreya [7K]
3 years ago
5

A 2kg object travels to the right at 5.5 m/s after colliding with a 3kg object, initially at rest, what are the speeds of each o

bject after the collision l?

Physics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

Velocity of Object with 2 kg= 3.390 m/s

Velocity of Object with 3 kg= 3.404 m/s

Explanation:

From the picture, it can be seen that object B is initially at rest while object A is travelling at a speed of 5m/s. After the collision, Object A moves at an angle of 65 degrees while object B moves at an angle of 37 degrees.

We also know that momentum of a closed system is conserved.

Initial momentum along the x-axis = 2*5.5 = 11

Initial momentum along y-axis = 0

Final momentum along x-axis= a*Cos(65)*2 +b*Cos(37) *3= 11 (a is the velocity of object A of 2 kg after collision where as b is the velocity of object B of 3 kg after collision. velocity is multiplied by cosines of the angle from x axis to give the horizontal component of the velocities).

Final momentum along y-axis = a*Sin(65)*2 - b*Sin(37)*3 =0 (We can see that vertical components of velocity are opposite in direction to each other)

Solve both the equations simultaneously for a and b.

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3 years ago
( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration
Talja [164]

Answer:

4.725 m/s^{2}

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then a=\frac {F}{m}  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula m=\frac {F}{a}

For m1= \frac {F}{13.5} and m2=\frac {F}{3.5} hence F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}

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On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The
Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

a = g (\sin  \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2

Velocity at the bottom

v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s

after travelling 4m , its velocity becomes 0

a=\frac{v^2-u^2}{2s}

a=\frac{0-u^2}{2s}

a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2

Coefficient of kinetic friction

μ = F/N

=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31

Therefore, the Coefficient of kinetic friction is 0.31

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snow_lady [41]

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Explanation:

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