A 9.7 kg block moving at 3 m/s makes a head-on elastic collision with a stationary block of mass 7.6 kg.
Find the velocity of the 7.6 kg block.
Answer in units of m/s.
9.7*3 + 7.6*0 = 9.7 Va + 7.6 Vb
so
Va = (29.1 -7.6Vb)/9.7
then
.5(9.7)(9)= .5(9.7)Va^2+.5(7.6)Vb^2
To find the velocity of the 7.6 kg block after the collision, we can use the principles of conservation of momentum and kinetic energy.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Where:
- m1 and m2 are the masses of the two blocks
- v1 and v2 are the initial velocities of the two blocks
- v1' and v2' are the final velocities of the two blocks
In this case, the first block has a mass of 9.7 kg and an initial velocity of 3 m/s. The second block has a mass of 7.6 kg and is initially stationary (so its initial velocity is 0 m/s). We want to find the final velocity of the second block, v2'.
Using the conservation of momentum equation, we have:
(9.7 kg * 3 m/s) + (7.6 kg * 0 m/s) = (9.7 kg * v1') + (7.6 kg * v2')
Simplifying the equation:
29.1 kg·m/s = 9.7 kg·v1' + 0 kg·m/s
Since the second block is initially at rest, the equation simplifies further:
29.1 kg·m/s = 9.7 kg·v1'
Now we can solve for v1':
Dividing both sides of the equation by 9.7 kg:
v1' = 29.1 kg·m/s / 9.7 kg
= 3 m/s
So, the final velocity of the 7.6 kg block after the collision is 3 m/s.