1) David has a mass of 86.4-kg and is skating with in-line skates. He sees his 24.8-kg younger brother up ahead standing on the sidewalk, with his back turned. Coming up from behind, he grabs his brother and rolls off at a speed of 2.99 m/s. Ignoring friction, find david's speed just before he grabbed his brother.
I do not know how to start this.
The law of conservation of linear momentum
m1•v = (m1+m2) •u
v= (m1+m2) •u/m1 =(86.4+24.8) •2.99/86.4 =3.85 m/s
To solve this problem, you can use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if no external forces act on it.
Step 1: Identify the variables given in the problem
- David's mass (m1) = 86.4 kg
- Younger brother's mass (m2) = 24.8 kg
- Initial speed of David and his brother (v1i) = ? (what you need to find)
- Final speed of David and his brother (v1f) = 2.99 m/s
- Final speed of the younger brother (v2f) = 0 m/s (as he is grabbed and rolls off with David)
Step 2: Apply the conservation of momentum equation
According to the conservation of momentum, the total initial momentum should equal the total final momentum.
Initial momentum = Final momentum
(m1 * v1i) + (m2 * 0) = (m1 * v1f) + (m2 * v2f)
Note: Since the younger brother is stationary initially (v2i = 0 m/s), his contribution to the initial momentum is zero.
Step 3: Solve the equation for the unknown variable
(m1 * v1i) + (m2 * 0) = (m1 * v1f) + (m2 * v2f)
(m1 * v1i) = (m1 * v1f)
v1i = v1f
In this case, David's speed just before grabbing his brother is the same as their final speed, which is 2.99 m/s.
Therefore, David's speed just before he grabbed his brother was 2.99 m/s.