A block with a mass of 2.5 kg on a frictionless track collides and sticks to a stationary block with a mass of 1.75 kg and the two more together at a velocity of 3.5 m/s. What was the velocity of the first block?
A. 2.5 m/s B. 7.5 m/s C. 6.0 m/s D. 0.6 m/s E. 0 m/s
pls show work ! thank u
To solve this problem, we can apply the law of conservation of momentum. According to this law, the total momentum before the collision should be equal to the total momentum after the collision, assuming no external forces are acting.
The momentum of an object can be calculated by multiplying its mass (m) by its velocity (v). So, the momentum before the collision can be expressed as:
Initial momentum (block 1) = m1 * v1
where m1 is the mass of the first block and v1 is its initial velocity.
After the collision, the two blocks stick together and move with a combined velocity of 3.5 m/s. Therefore, the momentum after the collision can be expressed as:
Final momentum (combined blocks) = (m1 + m2) * v2
where m2 is the mass of the second block and v2 is the final velocity after collision.
Since the total momentum is conserved, we can equate the initial and final momentum equations to find the initial velocity (v1):
m1 * v1 = (m1 + m2) * v2
Substituting the given values:
2.5 kg * v1 = (2.5 kg + 1.75 kg) * 3.5 m/s
Simplifying the equation:
2.5 kg * v1 = 4.25 kg * 3.5 m/s
Dividing both sides of the equation by 2.5 kg:
v1 = 4.25 kg * 3.5 m/s / 2.5 kg
v1 = 14.875 m/s / 2.5 kg
v1 = 5.95 m/s
Therefore, the velocity of the first block before the collision was approximately 5.95 m/s.
The closest answer choice from the given options is C. 6.0 m/s.