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kaheart
2 months ago
6

A ball collides with a vertical, unmovable wall. There is no friction between the wall and the ball (the only force acting on th

e ball during the collision is the normal force from the wall). ball colliding with wall 1) What is the direction of the x-component of the impulse imparted to the ball during the collision?
Physics
1 answer:
Yuliya22 [3.3K]2 months ago
7 0

Answer: in the same direction, meaning to the left.

Explanation:

The force acting perpendicularly to the surface in contact is such that it halts relative motion and, in the case of an elastic collision like this, restores the system's kinetic energy to its previous level. Thus, when a ball strikes an immovable surface, its speed (velocity magnitude) remains unchanged following the impact.

There will be a frictional force acting in parallel, but this should be examined separately for two main reasons:

The perpendicular force is constrained to the product of the coefficient of friction and the normal force. If this is insufficient to halt the ball, skidding will occur. The perpendicular force, which relies on the specific shape involved, does not align with the ball's center of mass, producing a torque that induces rotation. Once the ball rotates in such a way that the contact point is stationary, the momentum that would otherwise create friction force vanishes, causing the friction to cease, thus halting the ball's deceleration.

Consequently, the vertical component of the velocity will be inverted, whereas the horizontal component will be reduced to some extent, transferring a portion of kinetic energy into rotational energy. The rotation process will always negate the friction force before the horizontal velocity reaches zero, allowing the ball to keep moving in the same direction, albeit at a slower pace.

If instead you were to throw an elastic box (which wouldn't be able to rotate freely), it could indeed bounce back.

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inna [3103]

Let us denote a_1=3 m/s^2 as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is
F= ma_1
where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is
F=3m a_2
where a2 represents the new acceleration. 
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:
m a_1 = 3 m a_2
and solving for a2 gives us:
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5 0
4 months ago
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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [3030]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

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