A long rod is suspended by a frictionless bearing near its top. A lump of clay i

Question

A long rod is suspended by a frictionless bearing near its top. A lump of clay is projected from the right, collides with the block and sticks to it. Which one of the following statements is correct about the clay and rod system for this collision? A) The mechanical energy linear momentum and angular momentum remain constant. B The linear momentum and the angular momentum remain constant but the mechanical energy changes. C) Only the angular momentum remains constant. D) None of the quantities of mechanical energy, linear momentum, or angular momentum remain constant. Two objects are to fall straight downward under the of gravity with an acceleration g. Object is released from rest and object 2 is thrown downward from a somewhat greater height with an initial velocity v_. Graphs of their motion are shown above. The second object catches up with the fist object at time T. Which of the following expression correctly describes the initial difference in heights? A) T B) () T C) 1/2 () T D) 1/2 T^+ T A net force f is applied to a 2.0 kg object that is initially moving at 3.0 m/s. The Force varies as shown in the graph above. The speed of the object at the end of the time interval closest to which of the following? A) 5.0 m/s B) 7.0 m/s C) m/s D) 14 m/s

Explanation / Answer

9. (A)

because hinged is frictionless hence no non-conservative forces acting on the system hence mechanical energy,linear momentum and angular momentum reamina constant during all the time.

10.(A)

both particles having the same acceleration hence distance between them coverd by relative velocity which is constant and equal to v2-v1=v2 because v1 is zero.

11.(B)

area under force vs time graph gives the change in momentum. integral of F.dt= integral of m.dv hence area under the curve = m(v2-v1).

12 data in picture is not visible clearly.

but it was a problem of simple pendulum which oscilltion time period is given by

T=2*pi*sqrt(L/g)

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