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2 years ago

11

g A ball is thrown against the wall and bounces back with the same velocity. What type of collision is this

1 answer:

Softa [21]2 years ago

8 0

Answer:

Perfectly elastic collision

Explanation:

In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.

We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.

In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.

As a result of this, the collision is perfectly elastic

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Three beads are placed along a thin rod. The first bead, of mass m1 = 28 g, is placed a distance d1 = 1.5 cm from the left end o

Vladimir79 [104]

Answer:

Part a)

Center of mass with respect to the left end is given as

$r_{cm} = 5.63 cm$

Part b)

Center of mass with respect to middle bead is

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Center of mass with respect to middle bead is

$r_{cm} = 1.63 cm$

Explanation:

Part a)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = 1.5 cm$

$r_2 = 1.5 + 2.5 = 4 cm$

$r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm$

Now we have

$r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}$

$r_{cm} = 5.63 cm$

Part b)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = -d_2 = -2.5cm$

$r_2 = 0$

$r_3 = d_3 = 4.6 cm$

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Now plug in the values in above formula

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

$r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}$

$r_{cm} = 1.63 cm$

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