1.An elastic wire extend by 1.0cm when a load of 20g hangs from it.what additional load will be required to cause a further extension of 2.0cm?
2.A ball A of mass 0.4kg moving with velocity 5ms-1 collides head on with a ball B of mass 0.2kg moving with velocity 2ms-1, after the collision,it move with the velocity 3ms-1 and B with velocity 6ms-1
(a) Show that collision obeys the law of conversation of momentum.
(b) Is the collision elastic or inelastic.?
Yes
I do not under stand the question
1. To solve this problem, we can use Hooke's Law, which states that the extension of an elastic material is directly proportional to the applied load.
First, let's find the spring constant (k) of the wire. We can use the formula:
k = F / x
Where F is the force applied (load) and x is the extension.
Given:
Load 1 (F1) = 20g = 0.02kg (Note: 1g = 0.001kg)
Extension 1 (x1) = 1.0cm = 0.01m
k1 = F1 / x1
= 0.02kg / 0.01m
= 2 N/m
Now, we can calculate the force required for the additional extension:
Extension 2 (x2) = 2.0cm = 0.02m
F2 = k1 * x2
= 2 N/m * 0.02m
= 0.04 N
Therefore, an additional load of 0.04 Newtons will be required to cause a further extension of 2.0cm.
2. (a) To show that the collision obeys the law of conservation of momentum, we need to calculate the initial momentum (before the collision) and the final momentum (after the collision) for both ball A and ball B. If the total momentum is conserved, then the law of conservation of momentum is valid.
Initial momentum of ball A (pA1) = mass of ball A * velocity of ball A
= 0.4kg * 5ms^-1
= 2 kgms^-1
Initial momentum of ball B (pB1) = mass of ball B * velocity of ball B
= 0.2kg * 2ms^-1
= 0.4 kgms^-1
Total initial momentum (p1) = pA1 + pB1
Final momentum of ball A (pA2) = mass of ball A * velocity of ball A after collision
= 0.4kg * 3ms^-1
= 1.2 kgms^-1
Final momentum of ball B (pB2) = mass of ball B * velocity of ball B after collision
= 0.2kg * 6ms^-1
= 1.2 kgms^-1
Total final momentum (p2) = pA2 + pB2
If p1 = p2, then the momentum is conserved, indicating that the collision obeys the law of conservation of momentum.
(b) To determine if the collision is elastic or inelastic, we need to consider whether there is any loss of kinetic energy during the collision. If there is no loss of kinetic energy, the collision is elastic. If there is a loss of kinetic energy, the collision is inelastic.
To check for kinetic energy change, we can calculate the initial kinetic energy (KE1) and the final kinetic energy (KE2) for both ball A and ball B.
Initial kinetic energy of ball A (KEA1) = 0.5 * mass of ball A * (velocity of ball A)^2
= 0.5 * 0.4kg * (5ms^-1)^2
= 5 J
Initial kinetic energy of ball B (KEB1) = 0.5 * mass of ball B * (velocity of ball B)^2
= 0.5 * 0.2kg * (2ms^-1)^2
= 0.4 J
Total initial kinetic energy (KE1) = KEA1 + KEB1
Final kinetic energy of ball A (KEA2) = 0.5 * mass of ball A * (velocity of ball A after collision)^2
= 0.5 * 0.4kg * (3ms^-1)^2
= 1.8 J
Final kinetic energy of ball B (KEB2) = 0.5 * mass of ball B * (velocity of ball B after collision)^2
= 0.5 * 0.2kg * (6ms^-1)^2
= 3.6 J
Total final kinetic energy (KE2) = KEA2 + KEB2
If KE1 = KE2, then there is no loss of kinetic energy, indicating an elastic collision. If KE1 > KE2, then there is a loss of kinetic energy, indicating an inelastic collision.
By comparing the calculated values of kinetic energy in this example, KE1 = 5 J and KE2 = 5.4 J, which means that the total kinetic energy increased after the collision. Therefore, the collision is an inelastic collision.
I recalculated the velocities after the collision :
V3 = 1/3 m/s = velocity of M1 after the collision.
V4 = 22/3 = 7 1/3 m/s = velocity of M2 after the collision.
a. Momentum before the collision:
M1*V1 + M2*V2 = 0.4*5 + 0.2*(-2) = 1.6 kg-m/s.
Momentum after the collision:
M1*V3 + M2*V4 = 0.4*1/3 + 0.2*(22/3) = 4.8/3 = 1.6 kg-m/s.
b. During an elastic collision, momentum AND KE are conserved.
KE before the collision:
0.5M1*V1^2 + 0.5M2*V2^2 = 0.2*5^2 + 0.1*(-2)^2 = 5 + 0.4 = 5.4 Joules.
KE after the collision:
0.5M1*V3^2 + 0.5M2*V4^2 = 0.2*(1/3)^2 + 0.1*(22/3)^2 = 5.4 Joules.
1. 20g/1cm = (20g+x)/2cm.
20/1 = (20+x)/2.
X = ?
2. Given:
M1 = 0.4 kg, V1 = 5 m/s.
M2 = 0.2 kg, V2 = -2 m/s.
V3 = 3m/s = velocity of M1 after collision.
V4 = 6m/s = velocity of M2 after collision.
a. Momentum before collision:
M1*V1 + M2*V2 = 0.4*5 + 0.2*(-2) = 1.6 kg-m/s.
Momentum after collision:
M1*V3 + M2*V4 = 0.4*3 + 0.2*6 = 2.4kg-m/s.
The Momentum before and after collision should be equal. but they are not
unless we assume the 2m/s is positive. But it should be negative, because
the balls are moving in the opposite direction. Please check the given data
for errors.