Two blocks of masses
m1 = 1.50 kg
m2 = 3.00 kg
are each released from rest at a height of
h = 4.40 m
on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)
(a) Determine the velocity of each block just before the collision.
(b) Determine the velocity of each block immediately after the collision.
(c) Determine the maximum heights to which
rise after the collision.
a) both masses are released from rest
by energy conservation
change in potential energy = change in kinetic energy
mgh = 1/2mv2
so v= sqrt(2gh) where sqrt = squar root
for mass1 m1= 1.5kg
V1 = sqrt(2*9.8*4.40) = 9.29 m/sec
for mass M2
V2 = sqrt(2*9.8*4.40) = 9.29 m/sec
b) elastic collision both conservation of momentum and kinetic energy are observed
so velocity of M1 after collision (V 3) = 5*9.29/3 = 15.83 m/sec
and velocity of M2 after collision (V 4) = 9.29/3 = 3.1 m/sec
c) again energy conservation
1/2 mv2 = mgh
h = v2 /2g
so mass 1 H1 = 15.832 /2*9.8 = 12.78 m
and mass 2 H2 = 3.12 / 2*9.8 = 0.49 meter
Get Answers For Free
Most questions answered within 1 hours.