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A block of mass m 5 kg slides with an initial speed vi = 5 m/s to the right toward a ramp of mass M = 20 kg. The ramp itself can also slide without friction on the surface and has a frictionless face that the block can slide upon. The block slides up the ramp so that its center of mass rises a maximum height h, then back down until it is separated from the ramp. Let velocities to the left be indicated by a negative
(a) Determine the speed vh of the block-ramp system at the instant the block reaches its maximum height on the ramp m/s
(b) Determine the height h to which the block's center of mass rises meters
(c) Determine the final velocities vf and vr of the block and ramp, respectively Final velocity of the block Final velocity of the ramp = xm/s X m/s

Respuesta :

Answer:

Part a)

[tex]v_{cm} = 1 m/s[/tex]

Part b)

[tex]h = 1.02 m[/tex]

Part c)

[tex]v_1 = 3 m/s[/tex]

[tex]v_2 = 2 m/s[/tex]

Explanation:

Part a)

When block reached to maximum height then block and ramp will move together.

so here we will have

[tex]v_{cm} = \frac{mv}{M + m}[/tex]

[tex]v_{cm} = \frac{5(5)}{5 + 20}[/tex]

[tex]v_{cm} = 1 m/s[/tex]

Part b)

Now we can use energy conservation to find the height

[tex]\frac{1}{2}mv^2 = \frac{1}{2}(M + m)v_{cm}^2 + mgh[/tex]

[tex]\frac{1}{2}(5)(5^2) = \frac{1}{2}(20 + 5)1^2 + 5(9.81)h[/tex]

[tex]h = 1.02 m[/tex]

Part c)

When block and ramp separate finally then we have

[tex]mv = -mv_1 + Mv_2[/tex]

[tex]5\times 5 = -5 v_1 + 20 v_2[/tex]

[tex]4v_2 - v_1 = 5[/tex]

also by energy equation

[tex]\frac{1]{2}(5)(5^2) = \frac{1}{2}(5)v_1^2 + \frac{1}{2}(20)v_2^2[/tex]

[tex]25 = v_1^2 + 4 v_2^2[/tex]

by solving above equations we have

[tex]v_1 = 3 m/s[/tex]

[tex]v_2 = 2 m/s[/tex]

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