Impulse/Velocity 4

Question

Two objects travel towards each other and undergo a perfectly inelastic interaction. Before they collide, object A has a mass of 682g and a velocity of 11m/s, while object B has a mass of 550g and a velocity of -2.6m/s. The interaction takes 0.19s from start to finish.

Beware: if both initial velocities are positive (or negative), that is saying that they are travelling in the same direction (for positive numbers, both are going to the right), however because their speeds are different, they are technically still "travelling towards each other)! The point is, don't make one negative when you do the calculations unless it was negative as written here/above!

What is the impulse by A on B?    unit   

What is the force by A on B?    unit   

What is the impulse by B on A?    unit   

What is the force by B on A?    unit   

Answer 1

Impulse is defined as the change in momentum of an object.

To figure out the change in momentum of an object, we first need to figure out the final velocity of each object.

Since it is a perfectly inelastic collision, this means the two objects stick together.

Momentum must be conserved so:

m_av_a + m_bv_b = (m_a + m_b)v_final

v_final = (m_av_a + m_bv_b) / (m_a + m_b)

v_final = (682g x 11 m/s + 550g x (-2.6 m/s)) / (682g + 550g)

v_final = (6072 g m/s) / (1232 g) = 4.93 m/s

a) The impulse by A on B is the change in B's momentum (p)

Δp_b = m_b (v_final - v_b) = 550g x (4.93 m/s - (-2.6 m/s)) = 550g x 7.5m/s = 4140 g m/s

The basic unit for impulse is N s or kg m/s. To convert to these units we divide the impulse by 1000 to change grams to kilograms.

Δp_b = 4140 g m/s = 4.14 Ns

b) Impulse = Force x (time of interaction)

Force of A on B = Impulse on B / (time of interaction)

Force of A on B = 4.14 Ns / 0.19s = 21.8 N

c) The impulse by B on A is the change in A's momentum (p). We could go through the same steps as for part a, however there is an easier way. Since momentum is conserved, the impulse of B on A must be equal and opposite to the impulse of A on B.

So, Δp_a = -4.14 Ns

d) The force of B on A is also just opposite and equal to the force of A on B (Newton's 3rd law). So, force of B on A = -21.8 N.

Answer 2

Equating total momentum before and after yields

MU + mu = (M + m)v, where M=0.682kg, U = 11m/s, m = 0.55kg, u = –2.6m/s.

The impulse by A on B is mv – mu,

the impulse by B on A is Mv – MU.

Given an impulse J in a time T = 0.19s, the average force exerted is J/T.