A small rubber ball is thrown at a heavier, larger basketball that is still. The small ball bounces off the basketball. Assume t
2 answers:
A. How does the force on the small ball compare to the force on the basketball?
Answer:
Force will be same in magnitude but opposite in sign
Explanation:
As per Newton's III law every action has equal and opposite reaction force.
So here when small ball apply force on large ball then the two forces will be equal in magnitude but opposite in direction
B. Compare the total momentum of the two balls before and after the collision?
Answer:
Total momentum of two balls will remains same before and after collision
Explanation:
As we know that there is no external force on the system of two bodies so the sum of momentum of two balls before and after collision must be same.
So we can say initial momentum of small ball = final momentum of small + big ball together.
C. The mass of the basketball is 600 grams and its velocity before the small ball hits is 0 m/s. The mass of the small ball is 100 grams and its velocity is +5 m/s before the collision and -4 m/s afterward. What is the velocity of the basketball after the collision?
Answer:
Explanation:
By momentum conservation equation we have
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Answer:
C 0.85 j/g*k
Explanation:
The specific heat capacity of a material is given by:
where
Q is the amount of heat supplied to the object
m is the mass of the object
is the increase in temperature of the object
For the object in this problem, we have
m = 117 g is the mass
Q = 1200 J is the heat supplied
is the increase in temperature
Substituting into the formula, we find the specific heat:
Answer:
a). M = 20.392 kg
b). am = 0.56 (block), aM = 0.28
(bucket)
Explanation:
a). We got N = mg cos θ,
f =
=
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
M = 20.392 kg
b). .............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
.....................(iv)
We got, N = mg cos θ
∴
................(v)
Mg - 2T = M
(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),
Using equation (iv), we get,
