A 0.24-kg object traveling rightward at 3.5 m/s collides head-on and elastically with a 0.40-kg object traveling leftward at 2.1 m/s. (a) what are their velocities after collision? (b) If the objects are in contact for 0.010 s during collision, what is the magnitude of the average forcethey exert on each other?
Use the conservation ofmomentum to solve for one veloicty in terms of the other. then put that into the conservation of energy equation.
Thanks
To solve this problem, we'll use the principles of conservation of momentum and elastic collision.
(a) Let's start with the conservation of momentum equation:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Where:
m1 = mass of the first object = 0.24 kg
v1 = initial velocity of the first object = 3.5 m/s
m2 = mass of the second object = 0.40 kg
v2 = initial velocity of the second object = -2.1 m/s
v1' = final velocity of the first object (to be determined)
v2' = final velocity of the second object (to be determined)
Plugging in the known values:
(0.24 kg * 3.5 m/s) + (0.40 kg * -2.1 m/s) = (0.24 kg * v1') + (0.40 kg * v2')
0.84 kg·m/s - 0.84 kg·m/s = 0.24 kg·v1' - 0.84 kg·v2'
Simplifying further:
0.24 kg·v1' - 0.84 kg·v2' = 0
Now we need to solve for v1' and v2', which gives us the final velocities after the collision.
Next, we'll use the equation for conservation of kinetic energy in an elastic collision:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Plugging in the known values:
(1/2) * 0.24 kg * (3.5 m/s)^2 + (1/2) * 0.40 kg * (-2.1 m/s)^2 = (1/2) * 0.24 kg * (v1')^2 + (1/2) * 0.40 kg * (v2')^2
0.735 J + 0.882 J = (0.12 kg) * (v1')^2 + (0.08 kg) * (v2')^2
1.617 J = 0.12 kg * (v1')^2 + 0.08 kg * (v2')^2
Now we have two equations:
0.24 kg·v1' - 0.84 kg·v2' = 0
0.12 kg·(v1')^2 + 0.08 kg·(v2')^2 = 1.617 J
We can solve these two simultaneous equations to find v1' and v2'. Let's rearrange the first equation to solve for v1':
0.24 kg·v1' = 0.84 kg·v2'
v1' = (0.84 kg * v2') / 0.24 kg
Now we substitute this expression for v1' into the second equation:
0.12 kg·((0.84 kg * v2') / 0.24 kg)^2 + 0.08 kg·(v2')^2 = 1.617 J
Simplifying this equation will give us a quadratic equation in terms of v2'. Solve it to find v2', and then substitute the value of v2' back into the equation for v1' to find the final answer for part (a).
(b) To find the magnitude of the average force exerted during the collision, we can use the formula:
F_avg = Δp / Δt
Where:
Δp = change in momentum
Δt = time of contact during the collision
The change in momentum can be given by:
Δp = m2 * v2' - m2 * v2
And plugging in the known values:
Δp = 0.40 kg * v2' - 0.40 kg * (-2.1 m/s)
Simplifying further:
Δp = 0.40 kg * (v2' + 2.1 m/s)
Finally, we can substitute this value of Δp into the formula for average force to find the magnitude.