Please help need answer asp

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

Misha Larkins [42]

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.

4 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)