a kitten of mass 0.60kg leaps at 30 degrees to the horizontal out of a toy truck of mass 1.25kg causing it to move over horizontal ground at 4.0m/s .At what speed did the kitten leap?
0=m₁v₁cosα + m₂v₂.
v₁= - m₂v₂/m₁ cosα=
= - 1.25•4/0.60•cos30 = - 9.62 m/s
To solve this problem, we can use conservation of momentum. The total momentum before the leap is equal to the total momentum after the leap.
Before the leap, the only moving object is the toy truck. After the leap, both the toy truck and the kitten are moving.
Let's assume the speed of the kitten after the leap is v_kitten, and the speed of the toy truck after the leap is v_truck.
The momentum before the leap is:
initial momentum = mass of truck * velocity of truck before the leap
The momentum after the leap is:
final momentum = (mass of truck + mass of kitten) * (velocity of truck after the leap + velocity of kitten after the leap)
We can write these two equations as:
(initial momentum) = (final momentum)
(mass of truck * velocity of truck before the leap) = (mass of truck + mass of kitten) * (velocity of truck after the leap + velocity of kitten after the leap)
Plugging in the given values:
Mass of truck = 1.25 kg
Mass of kitten = 0.60 kg
Velocity of truck before the leap = 0 m/s (it's at rest)
Velocity of truck after the leap = 4.0 m/s
1.25 kg * 0 m/s = (1.25 kg + 0.60 kg) * (4.0 m/s + v_kitten)
Now we solve for v_kitten:
v_kitten = (1.25 kg * 0 m/s) / (1.85 kg) - 4.0 m/s
v_kitten = -2.40 m/s
The negative sign indicates that the kitten leaped in the opposite direction of the truck's motion. Therefore, the speed at which the kitten leaped is 2.40 m/s.
To find the speed at which the kitten leaped, we can use the principles of conservation of momentum and conservation of energy.
First, let's consider the conservation of momentum. In this scenario, we have two objects: the kitten and the toy truck. The initial momentum of the system is zero since both objects were initially at rest. After the leap, the toy truck moves with a velocity of 4.0 m/s.
Let's denote the velocity at which the kitten leaped as v. Now, the momentum of the system after the leap is the sum of the momentum of the kitten and the momentum of the toy truck. Since the momentum is a vector quantity, we need to consider the components in the horizontal direction.
The momentum of the kitten in the horizontal direction is given by:
m1 * v * cos(theta),
where m1 is the mass of the kitten (0.60 kg) and theta is the angle at which the kitten leaped (30 degrees).
The momentum of the truck in the horizontal direction is given by:
m2 * v2,
where m2 is the mass of the truck (1.25 kg) and v2 is the velocity of the truck (4.0 m/s).
According to the conservation of momentum, the sum of these two momentum components should be zero:
m1 * v * cos(theta) + m2 * v2 = 0.
Solving the equation for v gives us the speed at which the kitten leaped.