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A 4.0 kg mass is released from rest at the top of a frictionless ramp 2.0 m above ground. It slides down the ramp to a level, horizontal portion of frictionless track and then collides with a stationary 8.0 kg mass. Immediately after the collision, the larger mass has a velocity v_l = +3.5 m/s (i.e., heading to the right).
The velocity v_s of the smaller mass just after the collision:

A) -0.74 m/s B) +0.74 m/s C) -2.6 m/s D) +2.6 m/s E) 0 m/s

Respuesta :

lublana

Answer:

(A)−0.74m/s

Explanation:

We are given that

Mass ,[tex]m_1[/tex]=4 kg

Distance, d=2 m

[tex]m_2=8 kg[/tex]

[tex]v_2=+3.5 m/s[/tex]

[tex]u_2=0[/tex]

We have to find the velocity of smaller mass just after the collision.

Initial velocity of smaller mass,u=0

[tex]u_1=\sqrt{2gd}=\sqrt{2\times 9.8\times 2}=6.26m/s[/tex]

Velocity of smaller mass at the bottom

According to law of conservation of momentum

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]4\times 6.26+0=4v_s+8\times 3.5[/tex]

[tex]4v_s=4\times 6.26-8\times 3.5[/tex]

[tex]v_s=\frac{4\times 6.26-8\times 3.5}{4}[/tex]

[tex]v_s=-0.74m/s[/tex]

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