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V125BC [204]
3 years ago
5

In a collision, a 15 kg object moving with a velocity of 3 m/s transfers some of its momentum to a 5 kg object. What would be th

e velocity of the 5 kg object after the collision if the 15 kg object is still moving at 1 m/s?
Physics
1 answer:
Misha Larkins [42]3 years ago
4 0

The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter p:

p=mv

The total momentum is the sum of the momentums. The initial situation is the following:

m_A=15,\quad v_A=3,\quad m_B=5,\quad v_B=0

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).

So, at the beginning, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 3+5\cdot 0=45

At the end, we have

m_A=15,\quad v_A=1,\quad m_B=5,\quad v_B=x

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)

After the impact, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 1+5\cdot x=15+5x

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for x, and you'll have the final velocity of the 5-kg object.

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by Helen Keller

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4 Once I knew the depth where no hope was, and darkness lay on the face of all things. Then

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3 years ago
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
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Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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