A small block of mass m1 = 0.520 kg is released from rest at the top of a fricti
ID: 1793842 • Letter: A
Question
A small block of mass m1 = 0.520 kg is released from rest at the top of a frictionless, curve-shaped wedge of mass m2 = 3.00 kg, which sits on a frictionless, horizontal surface as shown in Figure a. When the block leaves the wedge, its velocity is measured to be vf = 4.60 m/s to the right as shown in Figure b. (a) What is the velocity of the wedge after the block reaches the horizontal surface? There is no friction between the wedge and the horizontal surface so the wedge is free to move. m/s to the left (b) What is the height h of the wedge? m
Explanation / Answer
mass m1 = 0.52 kg
mass m2 = 3 kg
velocity v1 = 4.6 m/s
initial momentum Pi = 0 kg.m/s
a) from law of conservation of momentum ,
0 = m1v1 + m2v2
v2 = -(m1/m2)v1
= -(0.52 kg / 3 kg)(4.6 m/s)
= - 0.80 m/s
b) the speed of block jest before the collision is
u1 = 2gh
from law of conservation of energy ,
(1/2)m1u12 + (1/2)m2u22 = (1/2)m1v12 + (1/2)m2v22
(1/2)m1(2gh)+ 0 = (1/2)m1v12 + (1/2)m2v22
h = {m1v12 + m2v22}/ 2m1g
= {(0.52 kg)(4.6 m/s)2 + (3 kg)( - 0.80m/s)2} / (2)(0.52 kg)(9.8 m/s2)
= 1.268 m
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