Given data
*The mass of Bruce is m_1 = 45 kg
*The initial velocity of the Bruce is u_1 = 2 m/s
*The mass of the biff is m_2 = 90 kg
*The initial velocity of the Biff is u_2 = -7 m/s
*The final velocity of the first glider is v_2 = -1 m/s
According to the law of conservation of linear momentum, the total linear momentum of a system remains constant
Applying the law of conservation of momentum as

Substitute the known values in the above expression as

Hence, the speed of the bruce knock backwards is v_1 = -10 m/s
Answer:
Although this question is not complete, I would give a general solution to this kind of problems.
If y(t) describes the position of a body with time such that
y(t) = at^(n) + bt^(m) + C
Then
V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)
Explanation:
As an example supplies the position of a particle is given by
y(t) = 4t³- 3t² + 9
V(t) = 4x3t²- 3x2t¹
V(t) = d(t)/dt = 12t² - 6t.
Another example,
If y(t) = 15t³ - 2t² + 30t -80
V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.
Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.
One last example,
y(t) = 2t^6 -3t²
V(t) = d(t)/dt = 12t^5 - 6t
As we know that centripetal acceleration is given as

here we know that
linear speed = 35 m/s
here radius of path is same as length of arm
R = 0.7 m
now from above above formula we have


so acceleration is 1750 m/s/s
Explanation:
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No, momentum is conserved so:
momentum before=momentum after
it is C. 100 kg m/s