A block of mass m_1 = 2.40 kg slides down a 30.0^\\circ incline which is 3.60 {\

Question

A block of mass m_1 = 2.40 kg slides down a 30.0^circ incline which is 3.60 { m m} high. At the bottom, it strikes a block of mass m_2 = 6.10 kg which is at rest on a horizontal surface . (Assume a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored.

Determine the speeds of the two blocks after the collision.
Enter your answers numerically separated by a comma.

 

 

Determine how far back up the incline the smaller mass will go.

,  =

 

 

A block of mass m_1 = 2.40 kg slides down a 30.0 degree incline which is 3.60 m high. At the bottom, it strikes a block of mass m_2 = 6.10 kg which is at rest on a horizontal surface . (Assume a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored. Determine the speeds of the two blocks after the collision. Enter your answers numerically separated by a comma. v m, v M = Determine how far back up the incline the smaller mass will go. h' =

Explanation / Answer

First, we use potential energy to find the speed at the bottom.

PE = KE
mgh = ( 1/2)mv2

gh = ( 1/2)v2

v = ( 2gh )

v = [( 2)(9.8)(3.6 )] = 8.4 m/s

Now, lets use momentum. The formula for a perfectly elastic collison is as follows:

vm1f = [( m1 - m2)/( m1 + m2)]vm1i

vm1f = [( 2.4-6.1)/(2.4+6.1)]8.4

vm1f = -3.656 m/s.

The formula for the final speed of m2 is as follows:

vm2f = 2m1 / ( ma + mb )vm1i

vm2f = [2(2.4)/(2.4+6.1)]8.4

vm2f = 4.74 m/s

b) Now that we know the new speed of m1 after the collision, which is -3.656 ( or 3.656 backwards ) we use it in the potential energy equation:

KE = PE

( 1/2)mv2 = mgh

(1/2)v2 = gh

( 1/2)(3.656)2 = ( 9.8 )h

h = .682 m

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