Determine the impulse of the wall on a 10 kg block which is sliding at a velocity of v = 7 m/s, necessary to stop the block. The coefficient of kinetic friction between the block and the horizontal plane is
m_k = 0.2. Negelect the friction impulse acting on the block during the collision.

1) 31.2 km
2) 21.5 km
3) 22.5 km
4) 32.4 km

The accleration of the block can be obtained before the velocity of the block before it strikes the wall.
S F_y = ma_y;
N - 10(9.81) = 10(0)
N = 98.1 N
Similarly S F_x = ma_x;
-0.2(98.1) = -10a
a = 1.962 m/s²

Applying the equation v2 = vo2 + 2ac(s - so) yields
v² = 7² + 2 ( -1.962) [10 - 0]
v² = 49 - 39.24 = 9.76 (m/s)²
v = 3.124 m/s

For the impulse and momentum, we have that
m(vx)1 + S F_x dt = m (v_x)_{2
10 (3.124) - I = 10 (0)
I = 31.24 N-s}

So the impulse of the wall on the block necessary to stop the block is I = 31.24 N-s.