Question

A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the collision? Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.

Answer 1

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble $m_1=41gram=0.041kg$

Velocity of marble $v_1=2.30m/sec$

Its collides with other marble of mass 25 gram

So mass of other marble $m_2=25gram=0.025kg$

Second marble is at so $v_2=0m/sec$

We have to find the velocity of second marble

From momentum conservation we know that

$m_1v_1+m_2v_2=(m_!+m_2)v$, here v is common velocity of both marble after collision

So $0.041\times 2.30+0.025\times 0=(0.041+0.025)v$

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

Answer 2

The speed of each marble immediately after the collision = 1.43mls

Calculation of perfect elastic collisions

The formula used to solve for perfect collision of particles = m1v1 +m2v2 = (m1 + m2) VC

Where m1 = 41.0g

m2 = 25g

v1 = 2.3 m/s

V2= 0 m/s

Therefore, VC = ?

Substitute the following variables into the equation above.

41 × 2.3 + 25 ×0 = ( 41 + 25) VC

94.3. + 0 = 66 VC

make VC the subject of formula,

VC = 94.3/66

VC = 1.43mls

Therefore, the speed of each marble immediately after the collision = 1.43mls

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