a) A 0.05 kg mass is used to accelerate a cart. At t=1 second, the 0.5 kg cart i
ID: 1779708 • Letter: A
Question
a) A 0.05 kg mass is used to accelerate a cart. At t=1 second, the 0.5 kg cart is at x=0.5 m, at rest. At t=2.5 seconds, the cart is at 1.5 m, traveling at 1.3 ± 0.3 m/s. Calculate and compare the impulse with the total change in momentum of the cart and mass.
b) What would be the effect of air resistance and friction on the final velocity and momentum of the cart?
c) If a cart of mass 1.2 kg, traveling at 0.5 m/s, collides elastically with a cart of mass 0.7 kg, initially at rest, what is the final velocity of the first cart?
d) If the two carts above stick together after the collision, what is the final velocity of the first cart?
e) What is the final velocity of the first cart in (c) if the second cart has mass 1.8 kg?
f) What is the final velocity of the first cart in (d) if the second cart has mass 1.8 kg?
Explanation / Answer
Using impulse momentum theorem
Impulse = change in momentum
change in momentum is final momentum - initial momentum
dP = pf-pi
dp = m*(vf-vi)
vi is the initial velocity = 0 m/sec
vf = 1.3 + or - 0.3 m/sec
then dp = m*(vf-vi) = 0.5*(1.3+0.3-0) = 0.8 kg m/sec
when vf = 1.3-0.3 m/sec
then dp = m*(vf-vi) = 0.5*(1.3-0.3-0) = 0.5 kg m/sec
b) due to air resistance and friction ,final velocity of the cart decreases and hence momentum p = m*v
if v decreases then momentum p also decreases
c)clearly the collision is a elastic collision
let m1 = 1.2 m/sec and u1 = 0.5 m/sec
m2 = 0.7 kg and u2 = 0 m/sec
then
vf = (m1-m2)*u/(m1+m2) = (1.2-0.7)*0.5/(1.2+0.7) = 0.132 m/sec
d) if two objects are tohether after collision,then that collision is a perfectly inelastic collision
then using
final velocity of the bodies is V = (m1*u1)/(m1+m2) = (1.2*0.5)/(1.2+0.7) = 0.316 m/sec
post the remaing questions as a different or sepearte question dear
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